An Inequality in Reciprocals


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.

An Inequality in Reciprocals


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 $u,v,w\ge 0.$


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

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


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

computing which,

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


Nassim Nicholas Taleb has kindly provided the following amazing illustration:

illusstration of an approach to solving this inequality in reciprocals


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