3

I have tried to improve the inequality by AM-GM. Here is what I have done:
Since
$$\sum_{cyc}{xy}\ge3\sqrt[3]{x^2y^2z^2}$$ $$\Rightarrow \left(\sum_{cyc}{xy}\right)^2 \ge3xyz\sqrt[3]{xyz}$$ That means we got to prove$\sqrt[3]{xyz} \ge x+y+z$.
However that is false in the case $x=1$,$y=2$,$z=3$. Any help?

Michael Rozenberg
  • 194,933
Cong Tran An
  • 171

2 Answers2

4

Simplify the inequality to $$x^2y^2+y^2z^2+x^2z^2\ge xyz(x+y+z)$$ If some of $x,y,z$ equals $0$, then the inequality is obvious.

So assume $xyz\neq 0$ and divide by $x^2y^2z^2>0$ both sides to get $$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge \frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}$$ This is true by the power of Cauchy Schwarz $$\frac{1}{y}\cdot\frac{1}{z}+\frac{1}{z}\cdot\frac{1}{x}+\frac{1}{x}\cdot\frac{1}{y}\leq \sqrt{\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{x^2}}\cdot\sqrt{\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}}$$

Svetoslav
  • 5,205
  • 2
  • 16
  • 35
  • The last inequality you were trying to prove is the classical $a^2+b^2+c^2\ge ab+bc+ca$ in the form of $(a,b,c)=(1/x,1/y,1/z)$. There are many ways to prove it. E.g., it's equivalent to $\frac{1}{2}\left((a-b)^2+(b-c)^2+(c-a)^2\right)\ge 0$, which is true; or by the Rearrangement Inequality; or direct Muirhead's Inequality (which is AM-GM in disguise). – user236182 Feb 20 '16 at 20:31
  • @user236182 I know, just for completeness I showed it in one of all the possible ways. – Svetoslav Feb 20 '16 at 20:33
1

By the Rearrangement Inequality, $$ \begin{align} \left(\frac1x+\frac1y+\frac1z\right)^2 &=\frac1{x^2}+\frac1{y^2}+\frac1{z^2}+2\left(\frac1{xy}+\frac1{yz}+\frac1{zx}\right)\\ &\ge\frac1{xy}+\frac1{yz}+\frac1{zx}+2\left(\frac1{xy}+\frac1{yz}+\frac1{zx}\right)\\ &=3\left(\frac1{xy}+\frac1{yz}+\frac1{zx}\right)\tag{1} \end{align} $$ Multiply both sides of $(1)$ by $(xyz)^2$ to get $$ (xy+yz+zx)^2\ge3xyz(x+y+z)\tag{2} $$

robjohn
  • 345,667