Problem 11984 From the American Mathematical Monthly

Problem

Problem 11984 From the American Mathematical Monthly - problem

Solution 1

Let $a=b+c-a,$ $v=c+a-b,$ $w=a+b-c.$ Then $u,v,w\gt 0$ and, with $\displaystyle s=\frac{a+b+c}{2},$ $sr=\sqrt{s(s-a)(s-b)(s-c)},$ implying

$\displaystyle 2r=\sqrt{\frac{uvw}{u+v+w}}.$

Moreover $5184=2^6\cdot 3^4.$ Hence, the required inequality is equivalent to

$\displaystyle \left(\frac{v+w}{2}\right)^6+\left(\frac{w+u}{2}\right)^6+\left(\frac{u+v}{2}\right)^6\ge 3^4\frac{u^3v^3w^3}{(u+v+w)^3},$

or,

$\displaystyle \frac{\displaystyle \left(\frac{v+w}{2}\right)^6+\left(\frac{w+u}{2}\right)^6+\left(\frac{u+v}{2}\right)^6}{3}\cdot\left(\frac{u+v+w}{3}\right)^3\ge u^3v^3w^3.$

This is true because, by the AM-GM inequality,

$\displaystyle\begin{align} \frac{\displaystyle \left(\frac{v+w}{2}\right)^6+\left(\frac{w+u}{2}\right)^6+\left(\frac{u+v}{2}\right)^6}{3}&\ge\frac{(vw)^3+(wu)^3+(uv)^3}{3}\\ &\ge \sqrt[3]{(vw)^3\cdot (wu)^3\cdot (uv)^3}\\ &=u^2v^2w^2, \end{align}$

and $\displaystyle\left(\frac{u+v+w}{3}\right)^3\ge uvw.$

Solution 2

With $\displaystyle s=\frac{a+b+c}{2},$ $abc=4RS=4Rrs.$ We are going to use 1) Euler's inequality, $r\ge 2r,$ Mitrinović's inequality, $R\ge 3\sqrt{3}r.$ With the AM-GM inequality,

$\displaystyle \begin{align} a^6+b^6+c^6 &\ge 3\sqrt[3]{a^6b^6c^6}=3(abc)^2\\ &\ge 48r^2R^2s^2\ge 48r^2(2r)^2(3\sqrt{3}r)^2=5184r^6. \end{align}$

Solution 3

Using Mitrinović's inequality, $R\ge 3\sqrt{3}r$ and applying Jensen's inequality for the function $f(x)=x^6,$ convex on $\mathbb{R},$ obtain

$\displaystyle \frac{a^6+b^6+c^6}{3} \ge \left(\frac{a+b+c}{3}\right)^6\;\Rightarrow\\ \begin{align} a^6+b^6+c^6&\ge\frac{2^6s^6}{3^5}\ge\frac{2^6(3\sqrt{3}r)^6}{3^5}\\ &=2^63^4r^6=5184r^6. \end{align}$

Acknowledgment

The problem is by Dan Sitaru; Solution 1 is by Roberto Tauraso; Solutions 2 and 3 are by Marian Cucoaneş. I am grateful to Dan for mailing to me the problem with a solution.

 

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