# Olympiad Inequality which can be solved using Cauchy–Schwarz inequality

$$x,y,z$$ are positive real numbers, such that $$x+y+z=3$$ . prove that :

$$\sum_{\mathrm{cyc}}\frac{x}{x^3+y^2+z} \leq 1$$

I tried many things , but I don't think any of those are worth of mentioning. However, I know problem can be solves using Cauchy–Schwarz inequality.

Please, share your ideas. Thanks.

• Since this is contest-math, please indicate a source for the problem. Sep 21, 2021 at 17:13
• There is an answer here but it uses Holder's inequality. Read this as well. I think this is safely not a current contest question because of the similarities to the questions I've just presented. Sep 21, 2021 at 17:16
• Unfortunately , I don't know where the problem comes from. It was on IZHO TST in my city few years ago, but I don't think they came up with it. Sep 21, 2021 at 17:20
• @TeresaLisbon Thank you very much! Sep 21, 2021 at 17:21
• You are welcome! If you are able to write an answer using only AM-GM and using the hints given, it will be useful. Sep 21, 2021 at 17:22

## 1 Answer

By C-S $$\sum_{cyc}\frac{x}{x^3+y^2+z}=\sum_{cyc}\frac{x\left(\frac{1}{x}+1+z\right)}{(x^3+y^2+z)\left(\frac{1}{x}+1+z\right)}\leq\sum_{cyc}\frac{(1+x+xz)}{(x+y+z)^2}\leq1.$$ Can you end it now?