Let $x,y,z$ be positive real numbers such that $xyz=1$. Prove that$$\frac{1}{(x+1)^2+y^2+1}+\frac{1}{(y+1)^2+z^2+1}+\frac{1}{(z+1)^2+x^2+1}\le\frac{1}{2}$$
I cant’t see how any known inequality such as AM-GM or Cauchy-Schwarz can be applied to this, and usually olympiads only require one of those 2.
Suggestions and solutions would be appreciated.
Taken from the 2019 Pan African Maths Olympiad http://pamo-official.org/problemes/PAMO_2016_Problems_En.pdf
We apply AM-GM Inequality with $x^2 + y^2 \geq 2xy$. \begin{align*} &\frac{1}{(x+1)^2 + y^2 + 1} + \frac{1}{(y+1)^2 + z^2 + 1} + \frac{1}{(z+1)^2 + x^2 + 1} \\ &= \frac{1}{x^2 + 2x + y^2 + 2} + \frac{1}{y^2 + 2y + z^2 + 2} + \frac{1}{z^2 + 2z + x^2 + 2} \\ &\leq \frac{1}{2xy + 2x + 2} + \frac{1}{2yz + 2y + 2} + \frac{1}{2zx + 2z + 2} \\ &= \frac{z}{2xyz + 2xz + 2z} + \frac{xz}{2x^2yz + 2xyz + 2xz} + \frac{1}{2zx + 2z + 2} \\ &= \frac{z}{2 + 2xz + 2z} + \frac{xz}{2z + 2 + 2xz} + \frac{1}{2zx + 2z + 2} \\ &= \frac{xz + z + 1}{2xz + 2z + 2} \\ &= \frac{1}{2} \end{align*}