$x,y,z$ 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.
Best 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?