# An Inequality in Reciprocals

### Problem

Leo Giugiuc has kindly communicated to me the following problem, along with an elegant solution. The problem is due to Hung Nguyen Viet, Hanoi, Vietnam.

### Preliminaries

Introduce function $f:\,[0, \infty)^3\rightarrow\mathbb{R},\;$ with

\begin{align} f(u, v, w) = 4(u^3+v^3+w^2)&+12(u^2w+v^2u+w^2v)\\ &-15(u^2v+v^2w+w^2u)-3uvw. \end{align}

Since $f\;$ is cyclic and homogeneous polynomial of degree three, then, according to the Pham Kim Hung's theorem, $f(u,v,w)\ge 0\;$ for all $u,v,w\ge 0\;$ if and only if $f(1,1,1)\ge 0;$ and, for all $u\ge 0,\;$ $f(u, 1,0)\ge 0.\;$ This is indeed so in our case: $f(1,1,1)=0\ge 0\;$ and $f(u,1,0)=(u-2)^2(4u+1)\ge 0,\;$ for $u\ge 0.$

Thus, we conclude that

$\displaystyle \frac{4}{3}(u^3+v^3+w^2)+4(u^2w+v^2u+w^2v)\ge 5(u^2v+v^2w+w^2u)+uvw,$

for all $u,v,w\ge 0.$

### Solution

So, for $t\in (0,1],$

\displaystyle\begin{align} \frac{4}{3}(t^{3a}+t^{3b}+t^{3c})&+4(t^{a+2b}+t^{b+2c}+t^{c+2a})\\ &\ge 5(t^{2a+b}+t^{2b+c}+t^{2c+a})+t^{a+b+c}. \end{align}

Next,

$\displaystyle\int^1_0\frac{1}{t}\left[\frac{4}{3}(t^{3a}+t^{3b}+t^{3c})+4(t^{a+2b}+t^{b+2c}+t^{c+2a})\right]\\ \displaystyle\qquad\;\space\;\;\ge \int^1_0\frac{1}{t}\left[5(t^{2a+b}+t^{2b+c}+t^{2c+a})+t^{a+b+c}\right].$

computing which,

\displaystyle\begin{align}\frac{4}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&+4\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\\ &\ge 5\left(\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\right)+\frac{1}{a+b+c}. \end{align}

### Illustration

Nassim Nicholas Taleb has kindly provided the following amazing illustration:

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