# Wu's inequality

### Solution 1

Pick point $F\;$ and form triangles with $F\;$ as a common vertex, sides $\{x,y\},\;$ $\{y,z\},\;$ and $\{z,x\},\;$ and angle of $120^{\circ}\;$ between each pair:

Let $[X]\;$ denote the area of shape $X.\;$ By the sine formula for the area of triangle,

\displaystyle\begin{align} [\Delta ABC] &= [\Delta ABF]+[\Delta BCF]+[\Delta CAF]\\ &=\frac{1}{2}xy\sin 120^{\circ}+\frac{1}{2}yz\sin 120^{\circ}+\frac{1}{2}zx\sin 120^{\circ}\\ &=\frac{\sqrt{3}}{4}(xy+yz+zx). \end{align}

By a Bottema's theorem (see Geometric Inequalities),

$\displaystyle (abc)^2\ge \left(\frac{4}{\sqrt{3}}[\Delta ABC]\right)^3.$

But by the Law of Cosine,

\begin{align} c^2&=x^2+y^2-2\cos 120^{\circ}\\ &=x^2+xy+y^2,\\ a^2 &= y^2+yz+z^2,\\ b^2 &= z^2+zx+x^2, \end{align}

which combine into,

\displaystyle\begin{align} (abc)^2 &= (x^2+xy+y^2)(y^2+yz+z^2)(z^+zx+x^2)\\ &\ge (\frac{4}{\sqrt{3}}[\Delta ABC])^3\\ &=[(\frac{4}{\sqrt{3}}\frac{\sqrt{3}}{4}(xy+yz+zx)]^3\\ &=(xy+yz+zx)^3. \end{align}

### Solution 2

This proof follows in the footsteps of Solution 1, except for the reason for the inequality $\displaystyle (abc)^2\ge \left(\frac{4}{\sqrt{3}}[\Delta ABC]\right)^3.$ As a matter of fact, this inequality expresses the isoperimetric property of equilateral triangles.

### Solution 3

Denote the difference of the left-hand side and the right-hand side of the required inequality as, say, $P.\;$ Then

\displaystyle\begin{align} P &= x^4y^2+x^4z^2+x^2y^4+x^2z^4+y^4z^2+y^2z^4\\ &+ x^yz+xy^4z+xyz^4\\ &-x^3yz^2-x^3y^2z-x^2y^3z-x^2yz3-xy^3z^2-xy^2z^3-3x^2y^2z^2. \end{align}

Now, by the AM-GM inequality,

\displaystyle\begin{align} \frac{x^4y^2}{2}+\frac{x^2z^4}{2}&\ge x^3yz^2\\ \frac{x^4z^2}{2}+\frac{x^2y^4}{2}&\ge x^3y^2z\\ \frac{y^4x^2}{2}+\frac{y^2z^4}{2}&\ge xy^3z^2\\ \frac{y^4z^2}{2}+\frac{x^4y^2}{2}&\ge x^2y^3z\\ \frac{z^4x^2}{2}+\frac{y^4z^2}{2}&\ge xy^2z^3\\ \frac{z^4y^2}{2}+\frac{x^4z^2}{2}&\ge x^2yz^3\\ x^4yz+zy^4z+xyz^4 &\ge 3x^2y^2z^2. \end{align}

Adding everything up we see that $P\ge 0.$

### Solution 4

• $\displaystyle x^2+xy+y^2\ge\frac{3}{4}(x+y)^2,\;$ for $\displaystyle\frac{1}{4}(x-y)^2\ge 0,$
• $\displaystyle y^2+yz+z^2\ge\frac{3}{4}(y+z)^2,\;$ for $\displaystyle\frac{1}{4}(y-z)^2\ge 0,$
• $\displaystyle z^2+zx+x^2\ge\frac{3}{4}(z+x)^2,\;$ for $\displaystyle\frac{1}{4}(z-x)^2\ge 0.$

Hence,

$\displaystyle(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\ge\frac{27}{64}(x+y)^2(y+z)^2(z+x)^2.$

Now observe that $x(y-z)^2+y(z-x)^2+z(x-y)^2\ge 0\;$ is equivalent to

$\displaystyle(x+y)(y+z)(z+x)\ge\frac{8}{9}(x+y+z)(xy+yz+zx),$

implying

\displaystyle\begin{align} \frac{27}{64}(x+y)^2(y+z)^2(z+x)^2 &\ge \frac{1}{3}(x+y+z)^2(xy+yz+zx)^2\ge (xy+yz+zx)^3. \end{align}

### Acknowledgment

The problem above (from the Romanian Mathematical Magazine) has been shared on facebook by Dan Sitaru with comments added by Rovshan Pirkuliyev (Solution 1), Soumava Pal (Solution 2), and Soumava Chakraborty (Solution 3). Solution 4 is by Le Viet Hung. Illustrations are by Nassim Nicaulas Taleb (## 1, 3) and Gary Davis (# 2); Illustration 4 is by Gio Pireds.

It was Leo Giugiuc's observation that the above, combined with

$3(u^2-uv+v^2)\ge u^2+uv+v^2,$

leads to another proposition

Let $x,y,z\gt 0.\;$ Prove that

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

Copyright © 1996-2018 Alexander Bogomolny

69117276