Another Hard Inequality

Question

Find the minimum of

$$\frac{y^2+1}{x^2+z+1}+\frac{x^2+1}{y+z^2+1}+\frac{z^2+1}{x+y^2+1}$$

with $x,y,z>-1$.

The answer is 2, when $x=y=z=1$.

I am stuck. Here is what I have so far:

$$\frac{y^2+1}{x^2+z+1}+\frac{x^2+1}{y+z^2+1}+\frac{z^2+1}{x+y^2+ 1}=3-\left(\frac{x^2-y^2+z}{x^2+z+1}+\frac{-x^2+y+z^2}{y+z^2+1}+\frac{x+y^2- z^2}{x+y^2+1}\right)\geq 3-\left(\frac{x^2-y^2+z}{2 x+z}+\frac{-x^2+y+z^2}{y+2 z}+\frac{x+y^2-z^2}{x+2 y}\right)=-\left(\frac{x^2-y^2+z}{2 x+z}+\frac{-x^2+y+z^2}{y+2 z}+\frac{x+y^2-z^2}{x+2 y}\right)+2+1=\left(\frac{(x-y) \left(x^2-y^2+z\right)}{(2 x+z) (x+y+z)}+\frac{(x-z) \left(x^2-y-z^2\right)}{(y+2 z) (x+y+z)}+\frac{(y-z) \left(x+y^2-z^2\right)}{(x+2 y) (x+y+z)}\right)+2$$

Source: https://brilliant.org/problems/minimum-11/?ref_id=1411033 (I did not write this question)

Answer 1

For $x=y=z=1$ we'll get a value $2$.

We'll prove that it's a minimal value.

Indeed, by C-S $$\sum_{cyc}\frac{y^2+1}{x^2+z+1}=\sum_{cyc}\frac{(y^2+1)^2}{(y^2+1)(x^2+z+1)}\geq\frac{\left(\sum\limits_{cyc}(y^2+1)\right)^2}{\sum\limits_{cyc}(y^2+1)(x^2+z+1)}.$$ Thus, it remains to prove that $$\left(\sum\limits_{cyc}(y^2+1)\right)^2\geq2\sum\limits_{cyc}(y^2+1)(x^2+z+1)$$ or $$\sum_{cyc}(x^4+2x^2+1-2x^2y-2x)\geq0$$ or $$\sum_{cyc}((x^2-y)^2+(x-1)^2)\geq0.$$ Done!