### Problem

Dan Sitaru has kindly posted a problem of his at the CutTheKnotMath facebook page, and Leo Giugiuc added his solution (Solution 1) later on. Leo used an extremely powerful method they invented with Dan. The method has repeatedly proved useful in establishing various inequalities. It also worked for the problem at hand. This particular inequality admits an easier, though, less universal solution (Solution 2). This reminded me of "Shark" Dodson from O'Henry's "The Roads We Take" who after running from home came to a road fork. Some times later, he was telling his friend: "I came to a place one evenin' where the road forked and I didn't know which fork to take. I studied about it for half an hour, and then I took the left-hand." To which the friend replied, "Oh, I reckon you'd have ended up about the same."

Prove that, for $x,y,z\in(0,\infty)\;$ that satisfy $xyz=1\;$ the following inequality holds:

$x(x-3(y+z))^2+(3x-(y+z))^2(y+z)\ge 27.$

### Solution 1

Introduce $y+z=2s\;$ and $yz=p^2.\$ $s\;$ is positive by definition, $p\;$ is chosen to be positive. By the AM-GM inequality $s\ge p.\;$ Also, $\displaystyle x=\frac{1}{p^2}.$ In terms of $s\;$ and $p\;$ the inequality becomes

$\displaystyle \frac{(6sp^2-1)^2}{p^6}+\frac{2s(2sp^2-3)^2}{p^4}\ge 27.$

This ca be manipulated into

$f_{p}(s)=8s^3p^6+12s^2p^4+6sp^2-27p^6+1\ge 1.$

For any fixed $p,\;$ the function $f_{p}:\;[p,\infty)\rightarrow\mathbb{R} is strictly increasing, i.e.,$f_{p}(s)\ge f_{p}(p).$It will suffice to show that$f_{p}(p)\ge 0.\;$But$f_{p}(p)=8p^9-15p^6+6p^3+1=(p^3-1)^2(8p^3+1)\ge 0$which completes the proof. ### Solution 2 Successively,$\begin{align} x(x-3(y+z))^2&+(3x-(y+z))^2(y+z)\\ &=[x^3-6x^2(y+z)+9x(y+z)^2]\\ &\;\;\;+[9x^2(y+z)-6x(y+z)^2+(y+z)^3]\\ &=x^3+3x^2(y+z)+3x(y+z)^2+(y+z)^3\\ &=(x+y+z)^3\ge 27\sqrt[3]{xyz}=27, \end{align}\$

by the AM-GM inequality.