TITLE: Minimum of f(x,y,z)
QUESTION [7 upvotes]: 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

REPLY [10 votes]: 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
$$\frac{2(a+b+c)^2}{3ab+3bc+3ac}\ge 2\Leftrightarrow (a-b)^2+(a-c)^2+(b-c)^2\ge 0$$