Gaining further insight into AMM 11832

This post is aimed as a second solution (v2.0) to AMM problem 11832 proposed by Donald Kunth, U.S.A.

My first proof in a previous blog entry: AMM problem 11832 by D. Knuth

(I believe this new line of computation gains more insight into the cog and wheels (working principles) behind the creation of the identity, rather than my initial approach, which was designed for dealing with a specific case).

We begin with the a theorem with generating functions:

The following is an excerpt from K. N. Boyadzheiv's paper:

Theorem: Given a pair of sequence $\displaystyle (a_n)_{n \ge 0}$ and $\displaystyle (b_n)_{n\ge 0}$ that are Binomial Inverses of each other, that is they satisfy the relation:

$\displaystyle a_n = \sum\limits_{k=0}^{n} (-1)^{n-k}\binom{n}{k} b_k$ and $\displaystyle b_n = \sum\limits_{k=0}^{n} \binom{n}{k} a_k$

for each $n \ge 0$, then the corresponding generating functions with central binomial coefficient weight are related as follows:

If $\displaystyle f_{b}(z) = \sum\limits_{n=0}^{\infty} \binom{2n}{n}b_nz^n$ then, $\displaystyle f_a(z) = \sum\limits_{n=0}^{\infty} \binom{2n}{n}a_nz^n = \frac{f_b\left(\frac{z}{1+4z}\right)}{\sqrt{1+4z}}$

Proof:

Lemma: For complex number $\alpha$ and $|z|$ in suitable region of convergence, we have:

$$\displaystyle \sum\limits_{n=0}^{\infty} (-1)^n\binom{\alpha}{n}a_nz^n = (1+z)^{\alpha}\sum\limits_{n=0}^{\infty}(-1)^n\binom{\alpha}{n}\left(\frac{z}{1+z}\right)^n\left(\sum\limits_{k=0}^{n}\binom{n}{k}a_k\right)$$

which is a consequence of Euler's series transformation formula.

The special case $\displaystyle \alpha = -\frac{1}{2}$ is what we are after, $\displaystyle \binom{-1/2}{n} = \frac{1}{4^n}\binom{2n}{n}$ and making the substitution $z \mapsto 4z$ we get the desired result. [Q.E.D.]

For further details please refer to the original paper itself.

Back to the problem at hand:

We have the Binomial transformation formula: $\displaystyle \sum\limits_{k=1}^{n}(-1)^{n-k}\binom{n}{k}(H_{2k} - H_k) = (-1)^{n-1}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)$ which can be easily established. Hence, putting it into the machinery provided by the theorem, we get:

$$g(z) := \sum\limits_{n=1}^{\infty} \binom{2n}{n}\left(\frac{4^n}{2n\binom{2n}{n}} - \frac{1}{2n}\right)z^n = \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{4^nz^n}{n} - \frac{1}{2} \sum\limits_{n=1}^{\infty} \binom{2n}{n}\frac{z^n}{n}= -\frac{1}{2}\log (1-4z) + \log \left(\frac{1+\sqrt{1-4z}}{2}\right)$$

and as a consequence:

$$f(z) := \sum\limits_{n=1}^{\infty} (-1)^{n-1}\binom{2n}{n} (H_{2n} - H_n)z^n = \frac{g\left(\frac{z}{1+4z}\right)}{\sqrt{1+4z}} = \frac{1}{\sqrt{1+4z}}\log \left(\frac{1+\sqrt{1+4z}}{2}\right)$$

Making the change $z \mapsto -z$ in the above identity we get:

$\displaystyle \sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n} - H_n)z^n = \frac{-1}{\sqrt{1-4z}}\log \left(\frac{1+\sqrt{1-4z}}{2}\right)$

Dividing both sides with $z$ and integration leads to D. Knuth's identity, as we saw in my previous proof:

$$\boxed{\sum\limits_{n=1}^{\infty} \binom{2n}{n} (H_{2n-1}- H_n)\frac{z^n}{n} = \log^2 \left(\frac{1+\sqrt{1-4z}}{2}\right) = \log^2 (C(z))}$$

Alternating version of Au-Yeung series: $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2}$

Problem: Closed form of $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2}$

We have seen how to deal with the Au-Yeung Series in three different ways in Closed form of Enrico Au-Yeung Series. However alternating versions (A) of Euler Sums often turn out to be trickier to deal with than the (H) counterpart.

Solution: We begin with the key identity:
                           
$$\begin{align}\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^{(3)}}{k+1} &= \frac{1}{n+1}\sum_{k=0}^{n} (-1)^k \binom{n+1}{k+1}\sum\limits_{j=0}^{k}\frac{1}{(j+1)^3}\tag{1}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}\frac{1}{(j+1)^3} \sum\limits_{k=j}^{n}(-1)^k \binom{n+1}{k+1}\tag{2}\\ &= \frac{1}{n+1}\sum_{j=0}^{n}(-1)^{j} \binom{n}{j}\frac{1}{(j+1)^3}\tag{3} \\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \binom{n+1}{j+1}\frac{1}{(j+1)^2} \tag{4}\\ &= \frac{1}{(n+1)^2}\sum_{j=0}^{n}(-1)^{j} \frac{1}{(j+1)^2}\sum\limits_{k=j}^{n}\binom{k}{j}\tag{5}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \sum_{j=0}^{k} (-1)^j\binom{k}{j}\frac{1}{(j+1)^2} \tag{6}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}\sum_{j=0}^{k} (-1)^j\binom{k+1}{j+1}\frac{1}{j+1}\tag{7}\\&= \frac{1}{(n+1)^2}\sum_{k=0}^{n} \frac{1}{k+1}H_{k+1}\tag{8}\\ &= \frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2}\tag{9}\end{align}$$

Justifications:
(1) Used $\displaystyle \binom{n+1}{k+1} = \frac{n+1}{k+1}\binom{n}{k}$ and applied change in order of summation.
(2) Used the identity: $\displaystyle \sum\limits_{k=j}^{n} (-1)^k\binom{n+1}{k+1} = \sum\limits_{k=j}^{n-1} (-1)^k\left(\binom{n}{k}+\binom{n}{k+1}\right) + (-1)^{n}\binom{n+1}{n+1} = (-1)^j\binom{n}{j}$ via telescoping summation.
(3) Used identity from (1).
(4) Used the identity: $\displaystyle \binom{n+1}{j+1} =  \sum\limits_{k=j}^{n} \binom{k}{j}$
(5) Interchanged order of summation.
(6) Reused identity from (1).
(7) Used the identity: $\displaystyle \sum\limits_{j=0}^{n} (-1)^j\binom{n+1}{j+1}\frac{1}{j+1} = H_{n+1}$.
(8) Used the identity: $\displaystyle \sum\limits_{k=1}^{n} \frac{H_k}{k} = \frac{H_n^2 + H_n^{(2)}}{2}$.

More generally, for a sequence of numbers $\{a_k\}_{k\ge 0}$, we denote $\displaystyle b_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} a_k$ and the binomial inversion formula tell us that $\displaystyle a_n = \sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} b_k$. Denoting $\displaystyle s_n = \sum\limits_{k=0}^{n} a_k$ we have,

\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{s_k}{k+1} &= \frac{1}{n+1}\sum\limits_{k=0}^{n} (-1)^k \binom{n+1}{k+1} \sum\limits_{j=0}^{k} a_j \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \binom{n+1}{k+1} \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} a_j \sum\limits_{k=j}^{n} (-1)^k \left[\binom{n}{k} + \binom{n}{k+1}\right] \\&= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} a_j = \frac{b_n}{n+1}
\end{align*}

Let us denote, $\displaystyle c_{n,r} = \sum\limits_{j_1 = 1}^{n}\sum\limits_{j_2 = 1}^{j_1} \cdots \sum\limits_{j_r = 1}^{j_{r-1}} \frac{1}{j_1 \cdots j_r}$ then we have, $\displaystyle c_{n,r} = \sum\limits_{j=1}^{n} \frac{c_{j,r-1}}{j} = c_{n-1,r} + \frac{c_{n,r-1}}{n}$ and consequently,

\begin{align*}
\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} &= \sum\limits_{k=1}^{n} (-1)^{k-1} \left[\binom{n-1}{k-1} + \binom{n-1}{k} \right] c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1}  \left[c_{k-1,r} + \frac{c_{k,r-1}}{k}\right] + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k}  c_{k,r} \\&= \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n-1}{k-1}  c_{k-1,r} + \sum\limits_{k=1}^{n} (-1)^{k-1}\binom{n-1}{k-1}\frac{c_{k,r-1}}{k} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k}  c_{k,r} \\&= \sum\limits_{k=1}^{n-1} (-1)^{k} \binom{n-1}{k}  c_{k,r} + \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1} + \sum\limits_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k}  c_{k,r} \\&= \frac{1}{n}\sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k}c_{k,r-1}
\end{align*}

Therefore, by $r$ times application we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} c_{k,r} = \frac{1}{n^r}$ and from binomial inversion result we have, $\displaystyle \sum\limits_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k^r} = c_{n,r}$.

Then, using $\displaystyle a_k = \frac{1}{(k+1)^r}$ and denoting by classical notation $\displaystyle s_k = H_{k+1}^{(r)}$ we have,

\begin{align*}
\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(r)}}{k+1} &= \frac{1}{n+1}\sum\limits_{j=0}^{n} (-1)^j \binom{n}{j} \frac{1}{(j+1)^r} \\&= \frac{1}{(n+1)^2}\sum\limits_{j=1}^{n+1} (-1)^{j-1} \binom{n+1}{j} \frac{1}{j^{r-1}} = \frac{c_{n+1,r-1}}{(n+1)^2}
\end{align*}

Our particular case being $r=3$, we have, $$\sum\limits_{k=0}^{n} (-1)^k \binom{n}{k} \frac{H_{k+1}^{(3)}}{k+1} = \frac{c_{n+1,2}}{(n+1)^2} = \frac{H_{n+1}^2 + H_{n+1}^{(2)}}{2(n+1)^2}$$

The Binomial Inversion formula tells us:

$$\sum\limits_{k=0}^{n} (-1)^k\binom{n}{k}\frac{H_{k+1}^2+H_{k+1}^{(2)}}{2(k+1)^2} = \frac{H_{n+1}^{(3)}}{n+1}$$

Now at this point the Euler accelaration formula(or the Euler Series Transformation) tells us:

$$\sum\limits_{n=0}^{\infty} (-1)^{n}\frac{H_{n+1}^2+H_{n+1}^{(2)}}{2(n+1)^2} = \sum\limits_{n=0}^{\infty} \frac{H_{n+1}^{(3)}}{2^{n+1}(n+1)}$$

That is after reindexing the summation: $$\sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = 2\sum\limits_{n=1}^{\infty}\frac{H_n^{(3)}}{2^nn}$$

The series in the RHS can be evaluated by observing that $\displaystyle \sum_{n=1}^{\infty}H_n^{(3)}x^n = \frac{\operatorname{Li}_3(x)}{1-x}$

Hence, $$\displaystyle \sum_{n=1}^{\infty}\frac{H_n^{(3)}}{n}x^{n} = \operatorname{Li}_4(x) -\frac{1}{2}\operatorname{Li}_2^2(x) - \log (1-x) \operatorname{Li}_3(x)$$

That is at $x = \dfrac{1}{2}$ we have: $$\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^2}{n^2} + \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} =  2\sum\limits_{n=1}^{\infty} \frac{H_n^{(3)}}{2^{n}n} = 2\log 2 \operatorname{Li}_3\left(\frac{1}{2}\right) - \operatorname{Li}_2^2\left(\frac{1}{2}\right) + 2\operatorname{Li}_4\left(\frac{1}{2}\right)$$

Now it remains to deal with the alternating part: $\displaystyle \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2}$ which was derived by R. Sitaramachandrarao in this paper.

$$\displaystyle  \sum\limits_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2} = -\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$

An alternative approach of calculating the same is to observe that:

$$\begin{align} \sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^{(2)}}{n^2} &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4}+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_{n-1}^{(2)}}{n^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}\sum_{k=1}^{n-1}\frac1{k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=k+1}^{\infty}\frac{(-1)^{n-1}}{n^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n-1}}{(k+n)^2k^2}\\ &=\frac{7}{8}\zeta(4)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n-1}\left(\frac1{k^2n(n+k)}-\frac1{kn(k+n)^2}\right)\\ &=\frac{7}{8}\zeta(4)-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{k^2n(n+k)}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\ &=\frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}(-1)^{k+n}\left(\frac{1}{kn^2(n+k)}+\frac{1}{k^2n(n+k)}\right)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&= \frac{7}{8}\zeta(4)-\frac{1}{2}\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{n^2k^2}+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2}\\&=\frac{7}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}}{kn(k+n)^2} \\ &= \frac{21}{8}\zeta(4)-\frac{1}{8}\zeta^2(2)+2\sum_{k=1}^{\infty}\frac{(-1)^kH_k}{k^3}\end{align}$$

The later alternating series evaluates to: $$\sum_{k = 1}^{\infty} \frac{(-1)^k}{k^3}H_k = -\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3)$$

which can be computed by expressing it as a logarithmic integrals. User Chris'ssis had shown me in the past an interesting way of dealing with the integrals.

Update: We have the more general relation, $$\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{H_n}{n^3}x^n = -\frac{1}{2}\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} -\frac{1}{2} \zeta(2)\operatorname{Li}_2(x)+\frac{7}{8}\zeta(4)-\frac{1}{4}\operatorname{Li}_2^2(1-x)+\frac{1}{4}\zeta^2(2)+\operatorname{Li}_4(x)\\+\frac{1}{4}\log^2 x\log^2(1-x)+\frac{1}{2}\log x\log (1-x)\operatorname{Li}_2(1-x)+\zeta(3)\log x -\log x\operatorname{Li}_2(1-x)$$

where, we may write $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$.

I am too lazy to add a derivation for this. It's quite long and tedious for a blog post.

For those who are interested in similar logarithmic integrals can refer to Lewin's book Polylogarithms and associated functions pg. 310 formulae no. (10) to (12)). I don't have an elementary approach to this alternating Euler Sum.

Upper bound for Mahler's Measure

Statement: A polynomial $f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0 \in \mathbb{C}[x]$ that has roots $z_1, z_2, ... , z_n \in \mathbb{C},$ then:

$$\prod_{k = 1}^{n} \max (1,|z_k|) \leq \sqrt{1+|a_{n-1}|^2 + ... + |a_{0}|^2}$$

(The obvious approach is to use Jensen's Formula, but why I decided to share it is because of the surprisingly elementary proof which I was not aware of to begin with!)

Math.Se link: Inequality relating coefficients and roots of a complex polynomial

Proof: The usual proof with Jensen's Formula:

Suppose the roots of polynomial $f(z)$ are $z_1,z_2,\cdots,z_n$,

where, $|z_1| \ge |z_2| \ge \cdots \ge |z_m| > 1 \ge |z_{m+1}| \ge \cdots \ge |z_n|$

Let, $g(z) = z^nf\left(\frac{1}{z}\right) = 1 + a_{n-1}z + \dots + a_0z^n$

Then, $\{1/z_k\}_{1\le k \le m}$ are the zeros of $g$ in the disk $|z| \le r = 1-\epsilon < 1$, where, $\epsilon$ is chosen such that $g(re^{i\theta}) \neq 0$ for $\theta \in [0,2\pi]$.

Then, using Jensen's Formula ([a proof on Math.Se]) we have: $\displaystyle \log|r^m z_1\cdots z_m| = \frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,d\theta$

i.e.,$$|r^m z_1\cdots z_m| = \exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,d\theta\right)$$

Applying Jensen Inequality, $\displaystyle \exp\left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,d\theta\right) \le \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,d\theta$

followed by Cauchy-Schwarz Inequality yields, 

$\displaystyle \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,d\theta \le \frac{1}{2\pi}\left(\int_0^{2\pi}\,d\theta\int_0^{2\pi} |g(re^{i\theta})|^2\,d\theta\right)^{1/2} = \sqrt{1+|a_{n-1}|^2 + ... + |a_{0}|^2}$

Therefore, $\displaystyle |r^m z_1\cdots z_m| \le \sqrt{1+|a_{n-1}|^2 + ... + |a_{0}|^2}$, and letting $\epsilon \to 0$, $r \to 1$ gives the desired result.

A Second Approach:

We can also use induction.

The 2-norm of a polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0$ is defined as,

$\lVert f \rVert_2 := \sqrt{|a_n|^2+|a_{n-1}|^2 + ... + |a_{0}|^2}$

and, we have: 

Lemma: $\displaystyle \lVert (x-z)f(x) \rVert_2 = \lVert (\overline{z}x-1)f(x) \rVert_2$

$\displaystyle \begin{align}\lVert (x-z)f(x) \rVert_2^2 &= \sum\limits_{j=0}^{n+1} |a_{j-1} - za_j|^2 \textrm{ ,where, $a_{-1} = a_{n+1} = 0$ in notation} \\ &= \sum\limits_{j=0}^{n+1} (a_{j-1} - za_j)\overline{(a_{j-1} - za_j)} \\ &= (1+|z|^2)\lVert f(x) \rVert_2^2 - \sum\limits_{j=0}^{n+1} (a_{j-1}\overline{za_j} + za_j\overline{a_{j-1}}) \\ &= (1+|z|^2)\lVert f(x) \rVert_2^2 - \sum\limits_{j=0}^{n+1} (z\overline{a_{j-1}}a_{j} + \overline{z}a_{j-1}\overline{a_{j}}) \\ &= \sum\limits_{j=0}^{n+1}(\overline{z}a_{j-1} - a_j)(z\overline{a_{j-1}} - \overline{a_j}) \\ &= \sum\limits_{j=0}^{n+1}|\overline{z}a_{j-1} - a_j|^2 = \lVert (\overline{z}x-1)f(x) \rVert_2^2 \end{align}$

as before let us write the roots of the polynomial in the order, $|z_1| \ge |z_2| \ge \cdots \ge |z_m| > 1 \ge |z_{m+1}| \ge \cdots \ge |z_n|$ .

Consider the polynomial: $\displaystyle g(x) = \prod\limits_{j=1}^m (\overline{z_j}x-1)\prod\limits_{j=m+1}^n (x - z_j) = \sum\limits_{j=0}^{n}b_jx^j$,

where, $|b_n| = |z_1z_2\cdots z_m|$

Now, 

$\begin{align}|z_1z_2\cdots z_m| = |b_n| &\le \sqrt{|b_n|^2+\cdots+|b_0|^2} \\ & = \lVert g(x) \rVert_2 \\ &= \lVert \frac{g(x)}{\overline{z_1}x - 1}(x-z_1) \rVert_2 \\ &= \lVert \frac{g(x)}{\prod\limits_{j=1}^m (\overline{z_j}x-1)}\prod\limits_{j=1}^m(x-z_1) \rVert_2 [\textrm{ ,by repeated application of lemma}] \\ &= \lVert f \rVert_2 \end{align}$

Evaluation of an Interesting limit: #(53) form Wolfram Catalans Constant

To Display the LaTeX follow this link

Problem: The limit due to Glaisher (1877) referred as (53) in wolfram page of Catalan's Contant(G):

$\displaystyle exp\left(-\frac{1}{2}+\frac{2G}{\pi}\right) = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}}\prod\limits_{k=1}^{n}\frac{(4k-1)^{4k-1}}{(4k-3)^{4k-3}}$

Link: Math.Se

Solution: We start with the integral representation of Catalan's constant:

$\displaystyle G = \frac{1}{2}\int\limits_0^{\frac{\pi}{2}} \dfrac{x}{\sin x}\,dx = \frac{\pi^2}{2}\int\limits_{0}^{\frac{1}{2}}\frac{x}{\sin \pi x}\,dx$

and we have, from partial fraction expansion of $\displaystyle \frac{\pi}{\sin x}$ ,

$$\begin{align}\frac{\pi}{\sin \pi x} &= \frac{1}{x} - 2x\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^2 - x^2}\\

&= \sum\limits_{n=0}^{\infty} \int_0^1 (-1)^nt^{n-x}\,dt + (-1)^n t^{n-1+x} \,dt\\

&= \int_0^1 \sum\limits_{n=0}^{\infty} (-1)^nt^{n-x} \,dt + \int_0^1 \sum\limits_{n=0}^{\infty}(-1)^nt^{n-1+x} \,dt\\

&= \int_0^1 \frac{1}{t^x(1+t)}\,dt + \int_0^1 \frac{1}{t^{1-x}(1+t)} \\

&=\int_0^1 \frac{1}{t^x(1+t)}\,dt +\int_1^{\infty} \frac{1}{t^x(1+t)}\,dt \\

&= \int_0^{\infty} \frac{1}{t^x(1+t)}\,dt\end{align}$$

Then, $\displaystyle G = \frac{\pi}{2}\int\limits_{0}^{\frac{1}{2}} \int_0^{\infty} \frac{x}{t^x(1+t)}\,dt \,dx = \frac{\pi}{2}\int_0^{\infty} \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx\,dt$

Now, $\displaystyle \int\limits_{0}^{\frac{1}{2}} \frac{x}{t^x(1+t)}\,dx = \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)$

Therefore, $\displaystyle \frac{2G}{\pi} = I = \int_0^{\infty} \frac{1}{t+1}\left(\frac{1}{\log^2 t} - \frac{1}{\sqrt{t}\log^ t} - \frac{1}{2\sqrt{t}\log t} \right)\,dt$

Making the substitiution, $t = e^{-w}$, the above integral becomes:

$$\begin{align} I &= \int_{-\infty}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw\\

&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(e^{-w} - e^{-w/2} + \frac{w}{2}e^{-w/2}\right)\,dw+\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1 - e^{-w/2} - \frac{w}{2}e^{-w/2}\right)\,dw\\

&=\int_{0}^{\infty} \frac{1}{w^2(1+e^{-w})}\left(1+ e^{-w} - 2e^{-w/2}\right)\,dw\end{align}$$

Introducing regulating factor $s$, we write:

$$I(s) = \int_0^{\infty} \frac{w^{s-2}}{1+e^{-w}}\left( 1+e^{-w} -2e^{-w/2} \right)\,dw$$

Therefore, $$\begin{align} I(s) &= \sum\limits_{n=0}^{\infty} (-1)^n \int_{0}^{\infty}\left( 1+e^{-w} -2e^{-w/2} \right)w^{s-2}e^{-nw}\,dw\\

&=\sum\limits_{n=0}^{\infty} (-1)^n\left(\frac{\Gamma(s-1)}{n^{s-1}} + \frac{\Gamma(s-1)}{(n+1)^{s-1}} -\frac{2^s\Gamma(s-1)}{(2n+1)^{s-1}}\right)\end{align}$$

Using the fact that, $\displaystyle \lim\limits_{s \to 0} s\Gamma(s-1) = -1$, we have,

$$I = \lim\limits_{s \to 0} \sum\limits_{n=0}^{\infty} \frac{(-1)^{n+1}}{s}\left(\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}\right)$$

Since, $$\begin{align} \lim_{s \to 0} \frac{\frac{1}{n^{s-1}} + \frac{1}{(n+1)^{s-1}} - \frac{2^s}{(2n+1)^{s-1}}}{s} &= n\left(2\log(n+1/2)-\log(n+1)-\log(n)\right) \\&+ \log(n+1/2)-\log(n+1)\end{align}$$

We have, $$\displaystyle \begin{align} I &= \sum_{n=0}^{\infty} (-1)^{n+1} n\left(2\log(n+1/2)-\log(n+1)-\log(n) \right) \\&+ \sum_{n=0}^{\infty} (-1)^{n+1} \left(\log(n+1/2)-\log(n+1)\right)\\

&= \frac{1}{2} + \lim\limits_{n \to \infty} \log \frac{1}{(4n+1)^{2n}} \prod_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}\end{align}$$

Thus, $$\displaystyle \exp{\frac{2G}{\pi} - \frac{1}{2}} = \lim\limits_{n \to \infty} \frac{1}{(4n+1)^{2n}} \prod\limits_{k=0}^{n-1}\frac{(4k+3)^{4k+3}}{(4k+1)^{4k+1}}$$

Adamchik's Integral : Omega Constant

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Problem: Adamchik's Integral representation of the Omega Constant.
$$\int\limits_{-\infty}^{\infty} \dfrac{1}{(e^x - x)^2 + \pi^2}\,dx = \dfrac{1}{1+\Omega}$$

LINKS: Omega Constant(Wiki) related to the Lambet-W-Function.
OEIS and Omega Constant(Wolfram)

Solution links:
(1) Robjohn's detailed solution on M.SE.
(2) WIKI-LINK with a proof (2.5)
(3) AOPS

Solution: Since, $ \displaystyle{\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}} = \frac{1}{{\left( {{e^z} - z - i\pi } \right)\left( {{e^z} - z + i\pi } \right)}} = \frac{1}{{2\pi i}}\left( {\frac{1}{{{e^z} - z - i\pi }} - \frac{1}{{{e^z} - z + i\pi }}} \right)}$

We have : $ \displaystyle\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}}dz} $

$ \displaystyle = \frac{1}{2\pi i}\left( {\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{e^z} - z - i\pi }}dz} - \int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{e^z} - z + i\pi }}dz} } \right)$

$ \displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{ - {e^{\left( {z + i\pi } \right)}} - \left( {z + i\pi } \right)}}dz} - \int\limits_{ - \infty }^{ + \infty } {\frac{1}{{ - {e^{\left( {z - i\pi } \right)}} - \left( {z - i\pi } \right)}}dz} } \right)$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty + i\pi }^{ + \infty + i\pi } {\frac{1}{{ - {e^w} - w}}dw} - \int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{ - {e^w} - w}}dw} } \right)$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} - \int\limits_{ - \infty + i\pi }^{ + \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} } \right) $

$ \displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} + \int\limits_{ + \infty + i\pi }^{ - \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} } \right)$.

But, $ \displaystyle{\left| {\frac{1}{{{e^w} + w}}} \right| \geqslant \frac{1}{{\left| {{e^w}} \right| + \left| w \right|}}\xrightarrow{{w \to \infty }}0}$
so, $ \displaystyle{\int\limits_c {\frac{1}{{{e^w} + w}}dw} = \int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} + \int\limits_{ + \infty + i\pi }^{ - \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} }$

Where, $\displaystyle{c:}$ rectangles with vertices $ \displaystyle{A\left( { - M, - \pi } \right),\;B\left( {M, - \pi } \right),\;C\left( {M,\pi } \right)\;\& \;D\left( { - M,\pi } \right)}$ with $ \displaystyle{M \to + \infty }$ and a proper direction of rotation.

The function $ \displaystyle{f\left( w \right) = \frac{1}{{{e^w} + w}}}$ is meromorphic inside the rectangle $ \displaystyle{c}$ (with some poles).

We seek the poles of $ \displaystyle{f\left( w \right) = \frac{1}{{{e^w} + w}}}$ inside $ \displaystyle{c}$ , therefore the roots of the equation $ \displaystyle{{e^w} + w = 0}$ .

Now, $ \displaystyle {e^w} + w = 0 \Rightarrow {e^{x + iy}} + x + iy = 0 \Rightarrow {e^x}\cos y + x = 0$ and $\displaystyle {{e^x}\sin y + y = 0}$.

For $ \displaystyle{y = 0}$ unique real roots of the equation $ \displaystyle{{e^x} + x = 0}$.

For $\displaystyle{y \ne 0}$ we $\displaystyle{{e^x} = - \frac{y}{{\sin y}}}$ which is impossible for $\displaystyle{y \in \left[ { - \pi ,\;\pi } \right]}$ (inside the rectangle $ \displaystyle{c}$ ). So if $ \displaystyle{\rho :}$ are the roots of the equation $\displaystyle{{e^x} + x = 0}$,

we have, $ \displaystyle\int\limits_c {\frac{1}{{{e^w} + w}}dw} = 2\pi i \cdot {\text{Res}}\left( {\frac{1}{{{e^w} + w}},w = \rho } \right)$

$\displaystyle = 2\pi i \cdot \mathop {\lim }\limits_{w \to \rho } \frac{{w - \rho }}{{{e^w} + w}} = \frac{{2\pi i}}{{{e^\rho } + 1}}$,

so, $\displaystyle\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}}dz}$

$\displaystyle = \frac{1}{{2\pi i}}\left( {\int\limits_{ - \infty - i\pi }^{ + \infty - i\pi } {\frac{1}{{{e^w} + w}}dw} + \int\limits_{ + \infty + i\pi }^{ - \infty + i\pi } {\frac{1}{{{e^w} + w}}dw} } \right) $

$\displaystyle = \frac{1}{{2\pi i}}\int\limits_c {\frac{1}{{{e^w} + w}}dw} = \frac{1}{{{e^\rho } + 1}}$

But, $\displaystyle{e^\rho } + \rho = 0 \Rightarrow {e^\rho } = \left( { - \rho } \right) \Rightarrow \left( { - \rho } \right){e^{\left( { - \rho } \right)}} = 1$, that is $\displaystyle{ - \rho = \Omega }$ where, $\displaystyle{\Omega :}$ the roots of the equation $\displaystyle{x{e^x} = 1}$.

and finally, $\displaystyle\int\limits_{ - \infty }^{ + \infty } {\frac{1}{{{{\left( {{e^z} - z} \right)}^2} + {\pi ^2}}}dz} = \frac{1}{{{e^\rho } + 1}} = \frac{1}{{\left( { - \rho } \right) + 1}} = \frac{1}{{1 + \Omega }}$

GEOMETRIC INEQUALITY WITH SUM OF ANGLE-BISECTORS EXTENDED TO CIRCUMCIRCLE

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Problem: 
Here, $\Delta ABC$ is a triangle, circumscribed by the circle of radius $R$, and with inradius $r$. If $AM_1,BM_2$ and $CM_3$ are the three angle bisectors extended to the circumcircle to the points $M_1,M_2$ and $M_3$. Then:

$$8r+2R \le AM_1+BM_2+CM_3 \le 6R$$

Addendum: The upper bound $6R$ follows easily from the fact that the diameter is the longest chord of a circle. So, $AM_1+BM_2+CM_3 \le 2R+2R+2R = 6R$. However, the upper bounds can be slightly strengthened to $5R+2r$ and the Trigonometric inequality in concern is:

$\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \le \sum\limits_{cyc} \cos A + \sum\limits_{cyc} \sin \frac{A}{2}$.

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Source: Problem by Sawarnik on M.SE.

M.SE:  Awesome solution by Jack D’Aurizio : HERE

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Problem Background: The problem (in the form of trigonometric inequality:$\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \ge -3+4\sum\limits_{cyc}\cos A$) appeared in Crux Mathematicorum Journal (1981) as Problem 696 proposed by George Tsintsifas, Thessaloniki, Greece [source : Inequalities proposed in “Crux Mathematicorum”(from vol. 1, no. 1 to vol. 4, no. 2 known as “Eureka”)].  Unfortunately I do not have access to the old volumes of the journal. So, I don't know the solutions that were posted in the subsequent issues.

The improved upper bound (addendum) is Problem 628 in the Crux Mathematicorum Journal Proposed by Roland H. Eddy, Memorial University of Newfoundland.[source :“Eureka”]

There are many more trigonometric inequalities due to Jack Garfunkel, Flushing, N. Y in Eureka that are remarkable variations and  improvements on both the upper and lower bounds.

Crux Mathematicorum Problem 1083 [1987](Proposed by Jack Garfunkel) If $A, B, C$ are the angles of a triangle, then:

$\displaystyle \frac{2}{\sqrt{3}}\sum\limits_{cyc} \sin A \le\sum\limits_{cyc} \cos\frac{B-C}{2} \le\frac{2}{\sqrt{3}}\sum\limits_{cyc} \cos \frac{A}{2}$
The Problem 1292 in the same journal by the same author asks the stronger bound:

$\displaystyle \sum\limits_{cyc} \cos\frac{B-C}{2} \le \frac{1}{\sqrt{3}}\left(\sum\limits_{cyc} \sin A + \sum\limits_{cyc} \cos \frac{A}{2}\right)$

The Problem 795 (proposed by Jack Garfunkel) in the same journal: Given a triangle $\Delta ABC$, let $t_a , t_b , t_c$ be the lengths of its internal angle bisectors, and let $T_a , T_b , T_c$ be the lengths of these bisectors extended to the circumcircle of the triangle. Prove that $\displaystyle T_a + T_b + T_c \ge \frac{4}{3}(t_a + t_b + t_c)$

Remark: Problem 795 is a weaker form of Problem 1083. Since, $\displaystyle t_a^2 = \frac{4bc}{(b+c)^2}s(s-a) \le s(s-a)$.

Hence, $\displaystyle (t_a + t_b + t_c) \le \sqrt{3(t_a^2 + t_b^2 + t_c^2)} \le \sqrt{3(s(s-a)+s(s-b)+s(s-c))} = \sqrt{3}s$

$\displaystyle = \sqrt{3}R\sum\limits_{cyc} \sin A \le \frac{3}{4}2R\sum\limits_{cyc} \cos\frac{B-C}{2} = \frac{3}{4}(T_a + T_b + T_c)$.

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Solution (Problem 696) : The lower bound:

The quadrilaretal $ABM_1C$ is cyclic. Thus the Ptolemy Theorem says:

$AB.CM_1+AC.BM_1=BC.AM_1$, that is $AM_1=\dfrac{b+c}{a}.BM_1$

(Since, angle subtended by equal chords at the circumcircle are equal and vice-versa, $BM_1=CM_1$, as $\angle BAM_1=\angle CAM_1=\dfrac{A}{2}$).

Also, $BM_1=2R\sin\frac{A}{2}$.

Similarly, getting expressions for $BM_2$ and $CM_2$, we have:

$\displaystyle AM_1+BM_2+CM_3=\sum\limits_{cyc} \frac{b+c}{a}.\left(2R\sin\frac{A}{2}\right)=\sum\limits_{cyc} \frac{\sin B+\sin C}{\sin A}.\left(2R\sin\frac{A}{2}\right)$

$\displaystyle =2R\sum\limits_{cyc} \frac{\sin B+\sin C}{\sin A}.\sin\frac{A}{2}=2R\sum\limits_{cyc} \frac{2\sin \frac{B+C}{2}.\cos\frac{B-C}{2}}{2\sin \frac{A}{2}.\cos \frac{A}{2}}.\sin \frac{A}{2}$

$\displaystyle = 2R\sum\limits_{cyc}\cos\frac{B-C}{2}$.

We also have, $r=4R\prod\limits_{cyc}\sin\dfrac{A}{2}=R(-1+\sum\limits_{cyc}\cos A)$,

Therefore, $8r+2R = 2R(-3+4\sum\limits_{cyc}\cos A)$

Thus, the inequality in the question is, $\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \ge -3+4\sum\limits_{cyc}\cos A = 1 + \frac{4r}{R}$

Using the substitution, $a=x+y$, $b=y+z$ and $c=x+z:$

$\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \ge -3+4\sum\limits_{cyc}\cos A$

$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{\sin A + \sin B}{2\cos \frac{C}{2}} \ge \sum\limits_{cyc} \frac{2(b^2+c^2-a^2)}{bc} – 3$

$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{\frac{2\Delta}{bc} + \frac{2\Delta}{ac}}{2\sqrt{\frac{1+\frac{a^2+b^2-c^2}{2ab}}{2}}} \ge \sum\limits_{cyc} \frac{2ab^2+2ac^2-2a^3-abc}{abc}$

$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{2\Delta(a+b)}{2\sqrt{\frac{(a+b+c)(a+b-c)}{4ab}}} \ge \sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$

$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{4\Delta(a+b)\sqrt{ab}}{\sqrt{(a+b+c)(a+b-c)}} \ge 2\sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$

$\displaystyle \Leftrightarrow \sum\limits_{cyc} \frac{(a+b)\sqrt{ab}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{\sqrt{(a+b+c)(a+b-c)}} \ge 2\sum\limits_{cyc} (2ab^2+2ac^2-2a^3-abc)$

$\displaystyle \Leftrightarrow \sum\limits_{cyc} (a+b)\sqrt{ab(a+c-b)(b+c-a)} \ge 2\sum\limits_{cyc} 2ab^2+2ac^2-2a^3-abc$

$\displaystyle \Leftrightarrow 2\sum\limits_{cyc} (x+z+2y)\sqrt{xz(x+y)(z+y)} \ge 2\sum\limits_{cyc} (xz(x+z) + 6xyz)$

$$\displaystyle \Leftrightarrow \sum\limits_{cyc} (x+z+2y)\sqrt{xz(x+y)(z+y)} \ge \sum\limits_{cyc} (x^2z + xz^2 + 6xyz)$$

On the other hand we have the inequality $\sqrt{(x+y)(z+y)} \ge (y+\sqrt{xz})$ (By Squaring and applying AM-GM)

Thus it suffices to show $\displaystyle \sum\limits_{cyc} (x+z+2y)(y\sqrt{xz}+xz) \ge \sum\limits_{cyc} (x^2z + xz^2 + 6xyz)$

$\displaystyle \Leftrightarrow (\sum\limits_{cyc} xz(x+z)) + 6xyz + (\sum\limits_{cyc} xy\sqrt{xz}+yz\sqrt{xz}+2y^2\sqrt{xz})\ge \sum\limits_{cyc} (x^2z + xz^2) + 18xyz$

$\displaystyle \Leftrightarrow \sum\limits_{cyc} xy\sqrt{xz}+yz\sqrt{xz}+2y^2\sqrt{xz}\ge 12xyz$

Which is AM-GM Inequality with $12$-terms.

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The improved upper bound:(Problem 628)

We start from $\displaystyle AM_1+BM_2+CM_3 = 2R\sum\limits_{cyc} \cos \frac{B-C}{2}$.

Then, $\displaystyle\sum\limits_{cyc} \cos \frac{B-C}{2} =\sum\limits_{cyc} \left(\cos\frac{B+C}{2}+2\sin\frac{B}{2}\sin\frac{C}{2}\right) =\sum\limits_{cyc} \sin\frac{A}{2} + 2\sum\limits_{cyc}\sin\frac{B}{2}\sin\frac{C}{2}$

Using the Jensen Inequality, we have $\displaystyle \sum\limits_{cyc} \sin\frac{A}{2} \le 3\sin\frac{A+B+C}{6} = \frac{3}{2}$.

So, it suffices to prove that, $\displaystyle 2\sum\limits_{cyc}\sin\frac{B}{2}\sin\frac{C}{2} \le \sum\limits_{cyc} \cos A = 1+ 4\prod\limits_{cyc} \sin\frac{A}{2} = 1+\frac{r}{R}$

For any triangle $\Delta ABC$, wlog, we have either, $\displaystyle A \ge \frac{\pi}{3} \ge B \ge C$ or, $\displaystyle C \ge B \ge \frac{\pi}{3} \ge A$. In either case we have $\displaystyle \left(\sin\frac{B}{2} - \frac{1}{2}\right)\left(\sin\frac{C}{2} - \frac{1}{2}\right)$.

That is, $\displaystyle 4\sin\frac{B}{2}\sin\frac{C}{2} \ge 2\left(\sin\frac{B}{2}+\sin\frac{C}{2}\right) - 1$

$\displaystyle \iff 1+ 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \ge 2\sin\frac{A}{2}\left(\sin\frac{B}{2}+\sin\frac{C}{2}\right) + \left(1 -\sin \frac{A}{2}\right)$

Now, $\displaystyle 2\sin\frac{B}{2}\sin\frac{C}{2} \le 2\left(\frac{\sin\frac{B}{2}+\sin\frac{C}{2}}{2}\right)^2 \le 2\sin^2 \frac{\pi - A}{4} = 1 -\sin\frac{A}{2}$

Therefore, $\displaystyle  1+\frac{r}{R} = 1+ 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} \ge 2 \sum\limits_{cyc} \sin\frac{A}{2}\sin\frac{B}{2}$

Combining, we have the desired upper bound: $\displaystyle AM_1+BM_2+CM_3 \le 5R+2r$

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Solution (Problem 1083): Lower bound of the Jack Garfunkel Inequality:

We start with observing that without loss of generality, for any triangle $\Delta ABC$, either $A \ge \frac{\pi}{3} \ge B \ge C$ or, $C \ge B \ge \frac{\pi}{3} \ge A$. Thus either way we have:

$\displaystyle \left(\cos \frac{A}{2} - \frac{\sqrt{3}}{2}\right)\left(\cos \frac{B}{2} - \frac{\sqrt{3}}{2}\right)\le 0 \implies  4\prod\,\cos\frac{A}{2} \le 2\sqrt 3\cdot\left(\cos\frac{A}{2}+\cos\frac{B}{2}\right)\cos\frac{C}{2}-3\cos\frac{C}{2}$

and, $\displaystyle \left(\cos\frac{A}{2}-\frac{\sqrt 3}{2}\right)\left(\cos\frac{C}{2}-\frac{\sqrt 3}{2}\right)\le 0 \implies  4\prod\,\cos\frac{A}{2}\le 2\sqrt 3\cdot\left(\cos\frac{A}{2}+\cos\frac{C}{2}\right)\cos\frac{B}{2}-3\cos\frac{B}{2}$

Adding these together we have (1):, $\displaystyle 8\prod\limits_{cyc} \cos\frac{A}{2}\le 2\sqrt 3\cdot\left(\sum\limits_{cyc} \cos\frac{B}{2}\cos\frac{C}{2}\right)+2\sqrt 3\cos\frac{B}{2}\cos\frac{C}{2}-3\left(\cos\frac{B}{2}+\cos\frac{C}{2}\right)$

Since, $\displaystyle 4cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=\sin A+\sin B+\sin C$

and, $2\cos\frac{B}{2}\cos\frac{C}{2}=\cos\frac{B-C}{2}+\cos\frac{B+C}{2}\le 1+\sin\frac{A}{2}$

From $(1)$ we have, $\displaystyle 2\sum\limits_{cyc}\sin A \le 2\sqrt{3}\cdot\left(\sum\limits_{cyc}\cos\frac{B}{2}\cos\frac{C}{2}\right)+\sqrt{3}\cdot\left(1+\sin\frac{A}{2}\right) - 3\left(\cos\frac{B}{2}+\cos\frac{C}{2}\right)$

Thus it suffices to show that (2): $\displaystyle 2\sqrt{3}\cdot\left(\sum\limits_{cyc}\cos\frac{B}{2}\cos\frac{C}{2}\right)+\sqrt{3}\cdot\left(1+\sin\frac{A}{2}\right) - 3\left(\cos\frac{B}{2}+\cos\frac{C}{2}\right) \le \sqrt{3}\sum\limits_{cyc} \cos\frac{B-C}{2}$

Since, $\displaystyle\sum\limits_{cyc}\cos\frac{B-C}{2}=2\sum\limits_{cyc}\cos\frac{B}{2}\cos\frac{C}{2} - \sum\limits_{cyc}\sin\frac{A}{2}$

The above inequality can be rewritten as: $\displaystyle 1+2\sin\frac{A}{2} \le \sqrt 3\cdot\left(\cos\frac{B}{2}+\cos\frac{C}{2}\right) - \left(\sin\frac{B}{2}+\sin\frac{C}{2}\right)$

$\displaystyle \iff \cos \frac{\pi}{3}+\cos\frac{B+C}{2} \le \left(\cos \frac{\pi}{6} \cos\frac{B}{2} - \sin \frac{\pi}{6}\sin\frac{B}{2}\right)+\left(\cos \frac{\pi}{6}\cos\frac{C}{2}-\sin \frac{\pi}{6}\sin\frac{C}{2}\right)$

$\displaystyle \iff 2\cos\left(\frac{\pi}{6}+\frac{B+C}{4}\right)\cos\left(\frac{\pi}{6}-\frac{B+C}{4}\right) \le \cos\left(\frac{\pi}{6}+\frac{B}{2}\right)+\cos\left(\frac{\pi}{6}+\frac{C}{2}\right)$

$\displaystyle \iff \cos\left(\frac{\pi}{6}-\frac{B+C}{4}\right) \le \cos\frac{B-C}{4} \tag{3}$

Since, $\cos \theta$ is a decreasing on $[0,\frac{\pi}{2}]$ and an even function, both cases $\displaystyle B \ge \frac{\pi}{3}$ and $\displaystyle B \le \frac{\pi}{3}$ implies $(3)$.

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Solution (further improvement) : An alternative way of deriving the Trigonometric Inequality from a stronger bound: 

We will prove the stronger inequality $\sum\limits_{cyc} \cos(\frac{B-C}{2}) \ge 1+ \sqrt{\dfrac{r^2+s^2+2rR}{2R^2}}$,


**Lemma 1:** $\prod \cos(\frac{B-C}{2}) = \prod \left(\dfrac{b+c}{a}\sin(\frac{A}{2})\right)$

$= \dfrac{(a+b+c)(ab+bc+ca)-abc}{abc}\prod
\sin(\frac{A}{2}) = \dfrac{2s.(r^2+s^2+4rR) – 4rRs}{4rRs}.\dfrac{r}{4R}$

$ = \dfrac{r^2+s^2+2rR}{8R^2}$

**Lemma 2:** $\sum\limits_{cyc} \dfrac{1}{\cos(\frac{B-C}{2})} = \dfrac{4R^2}{r^2+s^2+2rR}.\left(\left(\sum\limits_{cyc} \cos(\frac{B-C}{2})\right)^2-1\right)-1$

Proof: We have (i) $\sum\limits_{cyc} \dfrac{1}{\cos^2(\frac{B-C}{2})} = \dfrac{s^2+(r+4R)^2}{s^2}$

(ii) $\sum\limits_{cyc} \tan(\frac{A}{2}) = \dfrac{4R+r}{s}$

(iii) $\sum\limits_{cyc} \sin^2(\frac{A}{2}) = 1-\dfrac{r}{2R}$

So, $\left(\sum\limits_{cyc} \cos(\frac{B-C}{2})\right)^2 = \sum\limits_{cyc} \cos^2(\frac{B-C}{2}) + 2\sum\limits_{cyc} \cos(\frac{B-C}{2})\cos(\frac{C-A}{2}) $

$= \left(\sum\limits_{cyc} \dfrac{s}{2R}.\dfrac{1}{\cos(\frac{A}{2})} – \sin(\frac{A}{2}) \right)^2 + 2\prod \cos(\frac{B-C}{2}).\sum\limits_{cyc} \dfrac{1}{\cos(\frac{B-C}{2})} $

$= \dfrac{s^2}{4R^2}.\dfrac{s^2+(r+4R)^2}{s^2} – \dfrac{s}{R}.\dfrac{r+4R}{s} + (1-\dfrac{r}{2R}) + \dfrac{r^2+s^2+2rR}{4R^2}.\sum\limits_{cyc} \dfrac{1}{\cos(\frac{B-C}{2})}$

$= 1 + \dfrac{r^2+s^2+2rR}{4R^2}.\left(\sum\limits_{cyc} \dfrac{1}{\cos(\frac{B-C}{2})}\right)$,

as required.

Proof of the claim:

Starting from $\prod \left(\cos(\frac{B-C}{2}) – 1\right) \le 0$

$\iff 1 + \prod \cos(\frac{B-C}{2}).\sum\limits_{cyc} \dfrac{1}{\cos(\frac{B-C}{2})} \ge \sum\limits_{cyc} \cos(\frac{B-C}{2}) + \prod \cos(\frac{B-C}{2})$

$\iff 1 + \dfrac{r^2+s^2+2rR}{8R^2}.\left(\sum\limits_{cyc} \dfrac{1}{\cos(\frac{B-C}{2})} – 1\right) \ge \sum\limits_{cyc} \cos(\frac{B-C}{2})$

$\iff 1 + \dfrac{1}{2} \left[\left(\sum\limits_{cyc} \cos(\frac{B-C}{2})\right)^2 - 1\right] – \sum\limits_{cyc} \cos(\frac{B-C}{2}) \ge \dfrac{r^2+s^2+2rR}{4R^2}$

$\iff \sum\limits_{cyc} \cos(\frac{B-C}{2}) \ge 1 + \sqrt{\dfrac{r^2+s^2+2rR}{2R^2}}$.

Proving our claim.

Now, the part $1 + \sqrt{\dfrac{r^2+s^2+2rR}{2R^2}} \ge 1 + \dfrac{4r}{R}$ is simple.

It reduces to proving $s \ge 3\sqrt3 r$,

Now, $\dfrac{s}{3} = \dfrac{(s-a)+(s-b)+(s-c)}{3} \ge \{(s-a)(s-b)(s-c)\}^{1/3} = \{\dfrac{\Delta^2}{s}\}^{1/3} = (r^2s)^{1/3}$

**QED.**