An Inequality from Tibet

Problem

The following problem, due to Liu-Bao Qian, Lhasa Information Center, Tibet, has been posted at the CutTheKnotMath faceboook page by Leo Giugiuc, along with a beautiful solution by Claudia Nanuti, Diana Trailescu, Dan Sitaru and Leo Giugiuc.

For $a,b,c,d,e\gt 0,$ prove that

$\begin{align} (bcde+acde+abde&+abce+abcd)^4\\& \ge 125(a+b+c+d+e)(abcde)^3. \end{align}$

Solution 1

$\displaystyle\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right)^4\ge 125\left(\frac{1}{bcde}+\frac{1}{acde}+\frac{1}{abde}+\frac{1}{abce}+\frac{1}{abcd}\right).$

Make a substitution $\displaystyle\frac{1}{a}=y,$ $\displaystyle\frac{1}{b}=z,$ $\displaystyle\frac{1}{c}=u,$ $\displaystyle\frac{1}{d}=v,$ $\displaystyle\frac{1}{e}=w.$ Then $y,z,u,v,w\gt 0$ and we need to prove that

$(y+z+u+v+w)^4\ge 125(zuvw+yuvw+yzvw+yzuw+yzuv)$

Consider minimal monic polynomial $f(x)$ with roots $y,z,u,v,w:$

$f(x)=x^5-5sx^4+10qx^3-10tx^2+5rx-p,$

where $y+z+u+v+w=5s,$ $\displaystyle\sum yz=10q,$ $\displaystyle\sum yzu=10t,$ $\displaystyle\sum yzuv=5r,$ and $yzuvw=p.$

By Rolle's theorem, if $k,l,m,n$ are the roots of the derivative $f'(x),$ then $\min\{y,z,u,v,w\}\le\{k,l,m,n\}\le\max\{y,z,u,v,w\}.$ we have

$f'(x)=5(x^4-4sx^3+6qx^2-4tx+r).$

By the AM-GM inequality,

$k+l+m+n\ge 4\sqrt[4]{klmn},$

so that $4s\ge 4\sqrt[4]{r},$ i.e., $s^4\ge r,$ or, $625s^4\ge 625r,$ which is

$(y+z+u+v+w)^4\ge 125(zuvw+yuvw+yzvw+yzuw+yzuv),$

as required.

Solution 2

It was observed by Max Alekseyev that upon division by $(abcde)^4$ the inequality reduces to that of Maclaurin for the reciprocals.

 

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