A Cyclic Inequality in Triangle

Problem

A Cyclic Inequality in Triangle

Proof 1

By the AM-GM inequality,

$\displaystyle \frac{(a+b+c)^2}{4}\sum_{cycl}\frac{a^3(2s-a)}{b(2s-b)}\ge \frac{9\sqrt[3]{(abc)^2}}{4}\cdot 3\sqrt[3]{\prod_{cycl}a^2(2s-a)^2}.$

Thus, suffice it to show that

(A)

$\displaystyle \frac{9\sqrt[3]{(abc)^2}}{4}\cdot 3\sqrt[3]{\prod_{cycl}a^2(2s-a)^2}\ge 27a^2b^2c^2.$

Note that

(B)

$(2s-a)(2s-b)(2s-c)=(a+b)(b+c)(c+a)\ge 8abc.$

Combining (A) and (B) we get

$\displaystyle\begin{align} \frac{9\sqrt[3]{(abc)^2}}{4}\cdot 3\sqrt[3]{\prod_{cycl}a^2(2s-a)^2}&\ge \frac{9\sqrt[3]{(abc)^2}}{4}\cdot 3\sqrt[3]{(abc)^2\cdot 64(abc)^2}\\ &=27a^2b^2c^2. \end{align}$

Proof 2

By the AM-GM inequality,

$LHS \ge 3\sqrt[3]{a^2b^2c^2(a+b)^2(b+c)^2(c+a)^2}.$

Suffice it to show that

$\displaystyle\prod_{cycl}a^2(a+b)^2\ge\frac{729a^6b^6c^6}{s^6}.$

We have a sequence of equivalent statements:

$\displaystyle\begin{align} &\prod_{cycl}(a+b)\ge\frac{27(a+b+c)^2}{s^3}=\frac{432R^2r^2}{s},\\ &2abc+\prod_{cycl}ab(2s-c)\ge\frac{432R^2r^2}{s},\\ &2s^2\sum_{cycl}ab-4Rr^2\ge 432R^2r^2,\\ &s^2(s^2+4Rr+r^2)-2Rrs^2-216R^2r^2\ge 0. \end{align}$

To this we'll apply Gerretsen's inequality $s^2\ge 16Rr-5r^2:$

$\displaystyle\begin{align} s^2(s^2&+4Rr+r^2)-2Rrs^2-216R^2r^2\\&\ge (16Rr-5r^2)^2+(16Rr-5r^2)(2Rr+r^2)-216R^2r^2. \end{align}$

Suffice it to show that/p>

$\displaystyle(16Rr-5r^2)^2+(16Rr-5r^2)(2Rr+r^2)-216R^2r^2\ge 0.$

But this is equivalent to

$36R^2-77Rr+10r^2\ge 0,$

or, $(R-2r)(36R-5r)\ge 0,\;$ which is true due to Euler's inequality $R\ge 2r.$

Proof 3

By the AM-GM inequality,

$LHS \ge 3\sqrt[3]{a^2b^2c^2(a+b)^2(b+c)^2(c+a)^2}.$

Suffice it to show that

$\displaystyle\prod_{cycl}a^2(a+b)^2\ge\frac{3^3a^6b^6c^6}{s^6}.$

This is equivalent to

(*)

$\displaystyle (a+b+c)^3\prod_{cycl}(a+b)\ge 6^3a^2b^2c^2.$

In this form the inequality holds for $a,b,c\ge 0,\;$ not necessarily the sides of a triangle.

Now, $\displaystyle (a+b+c)^3 \ge 3^3abc,\;$ whereas $\displaystyle a+b\ge 2\sqrt{ab},\;$ $\displaystyle b+c\ge 2\sqrt{bc},\;$ $\displaystyle c+a\ge 2\sqrt{ca}.\;$ Multiplying the four gives (*).

Acknowledgment

Dan Sitaru has kindly posted the above problem from his book Math Accent, with two proofs - one (Proof 1) by Kevin Soto Palacios (Peru), the other (Proof 2) by Soumava Chakraborty (India), at the CutTheKnotMath facebook page.

 

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