Find the minimum of $P=\sum _{cyc}\frac{x+1}{1+y^2}$ under the constraints $x+y+z=3, x>0,y>0,z>0$

Question

For $x,y,z>0$ satisfying $x+y+z=3$, ind the minimum of $$P=\frac{x+1}{1+y^2}+\frac{y+1}{1+z^2}+\frac{z+1}{1+x^2}$$

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By AM-GM: $\frac{x+1}{1+y^2}=x+1-\frac{y^2\left(x+1\right)}{1+y^2}\ge x+1-\frac{y\left(x+1\right)}{2}=x-\frac{x}{2}+1-\frac{xy}{2}$

I think i need some method, the more the better

Answer 1

For $x=y=z=1$ we get a value $3$.

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

Indeed, we need to prove that $$\sum_{cyc}\frac{3+3x}{9+9y^2}\geq1$$ or $$\sum_{cyc}\frac{4x+y+z}{(x+y+z)^2+9y^2}\geq\frac{3}{x+y+z}$$ or $$\sum_{cyc}(7x^6+30x^5y+21x^5z-6x^4y^2+57x^4z^2+14x^3y^3+$$ $$+75x^4yz-6x^3y^2z+66x^3z^2y-258x^2y^2z)\geq0,$$ which is true because $\sum\limits_{cyc}x^6\geq\sum\limits_{cyc}x^4y^2$ is true by Rearrangement and the rest is true by AM-GM.

Done!