To Display the LaTeX follow this link
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:
$\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \le \sum\limits_{cyc} \cos A + \sum\limits_{cyc} \sin \frac{A}{2}$.
Source: Problem by Sawarnik on M.SE.
M.SE: Awesome solution by Jack D’Aurizio : HERE
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, $\displaystyle (t_a + t_b + t_c) \le \sqrt{3(t_a^2 + t_b^2 + t_c^2)} \le \sqrt{3(s(s-a)+s(s-b)+s(s-c))} = \sqrt{3}s$
$\displaystyle = \sqrt{3}R\sum\limits_{cyc} \sin A \le \frac{3}{4}2R\sum\limits_{cyc} \cos\frac{B-C}{2} = \frac{3}{4}(T_a + T_b + T_c)$.
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.
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: $\displaystyle AM_1+BM_2+CM_3 \le 5R+2r$
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 $(3)$.
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, $\dfrac{s}{3} = \dfrac{(s-a)+(s-b)+(s-c)}{3} \ge \{(s-a)(s-b)(s-c)\}^{1/3} = \{\dfrac{\Delta^2}{s}\}^{1/3} = (r^2s)^{1/3}$
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:
$\displaystyle \sum\limits_{cyc}\cos\frac{B-C}{2} \le \sum\limits_{cyc} \cos A + \sum\limits_{cyc} \sin \frac{A}{2}$.
Source: Problem by Sawarnik on M.SE.
M.SE: Awesome solution by Jack D’Aurizio : HERE
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, $\displaystyle (t_a + t_b + t_c) \le \sqrt{3(t_a^2 + t_b^2 + t_c^2)} \le \sqrt{3(s(s-a)+s(s-b)+s(s-c))} = \sqrt{3}s$
$\displaystyle = \sqrt{3}R\sum\limits_{cyc} \sin A \le \frac{3}{4}2R\sum\limits_{cyc} \cos\frac{B-C}{2} = \frac{3}{4}(T_a + T_b + T_c)$.
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.
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: $\displaystyle AM_1+BM_2+CM_3 \le 5R+2r$
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 $(3)$.
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.**
The imp is here, behold .... nice references though, how do you got them? Don't tell you have read all the 1980s issues of all those journals? :O :D
ReplyDelete