A Cyclic Inequality in Triangle II

Problem

A  Cyclic Inequality in Triangle II

Proof 1

WLOG, assume $a\ge b\ge c.\;$ Then $a^2\ge b^2\ge c^2\;$ but $\displaystyle\frac{1}{\sqrt{a}}\le \frac{1}{\sqrt{b}}\le \frac{1}{\sqrt{c}}.\;$ First employing the Rearrangement inequality, and then the AM-GM inequality,

$\displaystyle\begin{align} \sum_{cycl}\frac{a^2}{\sqrt{b}} &\ge \sum_{cycl}\frac{a^2}{\sqrt{a}}\\ &=a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}\\ &\ge 3\sqrt{abc}. \end{align}$

Thus,

(1)

$\displaystyle\sum_{cycl}\frac{a^2}{\sqrt{b}} \ge 3\sqrt{abc}.$

Using the Rearrangement inequality the second time and then the Chebyshev's inequality,

$\displaystyle\begin{align} \sum_{cycl}\frac{a^2}{\sqrt{b}} &\ge \sum_{cycl}\frac{a^2}{\sqrt{a}}\\ &=a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\\ &\gt \frac{1}{3}(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c}). \end{align}$

So that

(2)

$\displaystyle\sum_{cycl}\frac{a^2}{\sqrt{b}} \ge \frac{1}{3}\left(\sum_{cycl}a\right)\left(\sum_{cycl}\sqrt{a}\right).$

Multiplying (1) and(2), we get

(3)

$\displaystyle \sqrt{abc}\left(\sum_{cycl}\frac{a^2}{\sqrt{b}}\right)\ge (abc)(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c}).$

Define $x=a+b-c,\;$ $y=b+c-a,\;$ $z=c+a-b.\;$ Then $x+y\ge 2\sqrt{xy}\;$ and $b\ge\sqrt{xy}.\;$ Similarly, $a\ge\sqrt{xz}\;$ and $c\ge\sqrt{yz}.\;$ It follows that $abc\ge xyz.\;$ Thus we may continue (3):

$\displaystyle\begin{align} \sqrt{abc}\left(\sum_{cycl}\frac{a^2}{\sqrt{b}}\right)&\ge (abc)(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})\\ &\ge xyz(a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})\\ &=16S^2(\sqrt{a}+\sqrt{b}+\sqrt{c}), \end{align}.$

as required.

Proof 2

Since $(x+y)(y+z)(z+x)\ge 8 xyz,\;$ $abc\ge (a+b-c)(b+c-a)(c+a-b).\;$ Further,

$\displaystyle\begin{align} \sqrt{abc}\left(\sum_{cycl}\frac{a^2}{\sqrt{b}}\right)^2&\ge 3\sqrt[3]{\prod_{cycl}\frac{a^2}{\sqrt{b}}}\left(\sum_{cycl}\frac{a^2}{\sqrt{b}}\right)\sqrt{abc}\\ &=3abc\left(\sum_{cycl}\frac{a^2}{\sqrt{b}}\right)\\ &\ge \left(\prod_{cycl}(a+b-c)\right)\frac{3(a+b+c)^2}{\displaystyle\sum_{cycl}\sqrt{a}}\\ &\ge\left(\sum_{cycl}\sqrt{a}\right)(a+b+c) \left(\prod_{cycl}(a+b-c)\right)\\ &=(\sqrt{a}+\sqrt{b}+\sqrt{c})16S^2. \end{align}$

Acknowledgment

Dan Sitaru has kindly posted the above problem from his book Math Accent, with two proofs - one (Proof 1) by Soumava Pal, the other (Proof 2) by Diego Alvariz, at the CutTheKnotMath facebook page.

 

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