Hung Nguyen Viet's Inequality with Radicals and Chebyshev

Problem

Hung Nguyen Viet's Inequality with Radicals and Chebyshev

Proof 1

The required inequality is equivalent to $\displaystyle\sum_{cycl}a\sin\frac{A}{2}\ge\sum_{cycl}\sqrt{bc}\sin\frac{A}{2}.\,$ The sequences $a,\,$ $b,\,$ $c\,$ and $\displaystyle\sin\frac{A}{2},\,$ $\displaystyle\sin\frac{B}{2},\,$ $\displaystyle\sin\frac{C}{2}\,$ are similarly ordered. Hence, by Chebyshev's inequality,

$\displaystyle 3\sum_{cycl}a\sin\frac{A}{2}\ge (a+b+c)\left(\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\right).$

The sequences $\sqrt{bc},\,$ $\sqrt{ca},\,$ $\sqrt{ab}\,$ and $\sin\frac{A}{2},\,$ $\sin\frac{B}{2},\,$ $\sin\frac{C}{2}\,$ are oppositely ordered. Hence, by Chebyshev's inequality,

$\displaystyle \left(\sum_{cycl}\sqrt{bc}\right)\left(\sum_{cycl}\sin\frac{A}{2}\right) \ge 3\sum_{cycl}\sqrt{bc}\sin\frac{A}{2}.$

But, due to the Rearrangement inequality, $a+b+c\ge\sqrt{bc}+\sqrt{ca}+\sqrt{ab},\,$ completing the proof.

Proof 2

We have to prove $\displaystyle\sum_{cycl}a\sin\frac{A}{2}\ge\sum_{cycl}\sqrt{bc}\sin\frac{A}{2}.\,$ Now,

$\displaystyle\begin{align} \sum_{cycl}a\sin\frac{A}{2}&=\sum_{cycl}\frac{a\sqrt{(s-b)(s-c)}}{\sqrt{bc}}\\ &\ge 2\sum_{cycl}\frac{a}{b+c}\sqrt{(s-b)(s-c)}, \end{align}$

by the AM-GM inequality. Now sequences $(a,\,$ $b,\,$ $c)\,$ and $\displaystyle\left(\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+c}\right)\,$ are similarly ordered, while the the sequences $(\sqrt{bc},\,$ $\sqrt{ca},\,$ $\sqrt{ab})\,$ and $(\sqrt{(s-b)(s-c)},\,\sqrt{(s-c)(s-a)},\,\sqrt{(s-a)(s-b)})\,$ are oppositely ordered. To continue, using first Chebyshev's inequality and then Nesbitt's and the Rearrangement inequalities,

$\displaystyle\begin{align} \sum_{cycl}a\sin\frac{A}{2}&\ge\frac{2}{3}\left(\sum_{cycl}\frac{a}{b+c}\right)\left(\sum_{cycl}{(s-b)(s-c)}\right)\\ &\ge\frac{2}{3}\cdot\frac{3}{2}\sum_{cycl}\sqrt{(s-b)(s-c)}\\ &=\sum_{cycl}\sqrt{(s-b)(s-c)}\\ &\ge\sum_{cycl}\sqrt{bc}\ge\sum_{cycl}\sqrt{bc}\sin\frac{A}{2}.\\ \end{align}$

Acknowledgment

Leo Giugiuc has kindly posted at the CutTheKnotMath facebook page the above problem, due to Hung Nguyen Viet, with his solution (Proof 1). Proof 2 is by Soumava Chakraborty.

 

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