Inequality with $x^2+y^2+z^2=3$

Question

If x,y,z are positive real numbers and $x^2+y^2+z^2=3$ prove that $$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le\frac{1}{xyz}.$$ I have tried to prove it this way: $$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le\frac{1}{xyz}$$ $$\Leftrightarrow \sum_\text{cyc}\frac{xyz}{x^3+y^2+z}\le1$$ $$\Leftrightarrow \sum_\text{cyc}\frac{1}{\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}}\le1$$ But we know that $$\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}\ge3\sqrt[3]{\frac{x^2}{yz}\cdot\frac{y}{xz}\cdot\frac{1}{xy}}=\frac{3}{\sqrt[3]{z^2y}}$$ $$\Rightarrow \sum_\text{cyc}\frac{1}{\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}}\le\sum_{cyc}\frac{\sqrt[3]{z^2y}}{3}$$ We now have to prove that $\sum_\limits\text{cyc}\frac{\sqrt[3]{z^2y}}{3}\le1$. $$\frac{\sqrt[3]{z^2y}}{3}+\frac{\sqrt[3]{x^2z}}{3}+\frac{\sqrt[3]{y^2x}}{3}=\frac{\sqrt[3]{1\cdot z^2\cdot y}}{3}+\frac{\sqrt[3]{1\cdot x^2\cdot z}}{3}+\frac{\sqrt[3]{1\cdot y^2\cdot x}}{3}$$ $$\le\frac{\frac{1+ z^2+ y}{3}+\frac{1+ x^2+ z}{3}+\frac{1+ y^2+ x}{3}}{3}=\frac{3+3+x+y+z}{9}$$ And the last part: $\;\dfrac{3+3+x+y+z}{9}\le1 \Leftrightarrow x+y+z\le3$ or $$\frac{x+y+z}{3}\le \sqrt{\frac{x^2+y^2+z^2}{3}}$$ which is true. I hope that it's correct. Any suggestions are welcome.

Answer 1

I think your proof is right and very nice!

Also, we can use C-S:

$$\sum_{cyc}\frac{1}{x^3+y^2+z}=\sum_{cyc}\frac{x+y^2+z^3}{(x^3+y^2+z)(x+y^2+z^3)}\leq\sum_{cyc}\frac{x+y^2+z^3}{(x^2+y^2+z^2)^2}.$$

Answer 2

Alternative solution, by AM-GM

$$\frac{x^3+y^2+z}{3}\ge\sqrt[3]{x^3y^2z}\implies \frac{1}{x^3+y^2+z}\le\frac{1}{3\sqrt[3]{x^3y^2z}}=\frac{\sqrt[3]{z^2y}}{3xyz}$$

thus

$$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le \sum_\text{cyc} \frac{1}{3\sqrt[3]{x^3y^2z}}\le\sum_\text{cyc}\frac{\sqrt[3]{z^2y}}{3xyz} \le 1$$