Minimum of f(x,y,z)

let $x,y,z> -1$,then how can we find the minimum of this function $$f(x,y,z)=\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}$$ I think that we can use a famous inquality but I can not find it.

So thanks

For every $t\in\mathbb{R}$, we have $$t\le\frac{1+t^2}{2}$$ thus $$\frac{1+x^2}{1+y+z^2}\ge\frac{1+x^2}{1+z^2+\frac{1+y^2}{2}}$$ $$\frac{1+y^2}{1+z+x^2}\ge\frac{1+y^2}{1+x^2+\frac{1+z^2}{2}}$$ $$\frac{1+z^2}{1+x+y^2}\ge\frac{1+z^2}{1+y^2+\frac{1+x^2}{2}}$$ Set $a=1+x^2$ , $b=1+y^2$ and $c=1+z^2$, so $$f(x,y,z)\ge \frac{2a}{2c+b}+\frac{2b}{2a+c}+\frac{2c}{2b+a}$$ By application of Cauchy inequality, we have $$\left(\frac{2a}{2c+b}+\frac{2b}{2a+c}+\frac{2c}{2b+a}\right)\left[\,a(2c+b)+b(2a+c)+c(2b+a)\,\right]\ge 2(a+b+c)^2$$ thus $$\left(\frac{2a}{2c+b}+\frac{2b}{2a+c}+\frac{2c}{2b+a}\right)\ge \frac{2(a+b+c)^2}{3ab+3bc+3ac}\ge 2$$ finally $$\color{red}{f(x,y,z)\ge 2}$$

Note

If $u=\left(\sqrt{\frac{a}{2c+b}}\ ,\, \sqrt{\frac{b}{2a+c}}\, ,\,\sqrt{\frac{c}{2b+a}}\right)$ and $v=\left(\sqrt{a(2c+b)}\,,\,\sqrt{b(2a+c)}\,,\,\sqrt{c(2b+a)}\,\right)$ then $$\|u\|^2\,\|v\|^2\ge(u.v)^2$$ and $$\frac{2(a+b+c)^2}{3ab+3bc+3ac}\ge 2\Leftrightarrow (a-b)^2+(a-c)^2+(b-c)^2\ge 0$$