Find a minium value of a function Given x,y,z are positive real numbers such that
$$
x^2+y^2+6z^2=4z(x+y).
$$
Find the minimum value of the following function
$$
P=\frac{x^3}{y(x+z)^2}+\frac{y^3}{x(y+z)^2}+\frac{\sqrt{x^2+y^2}}{z}
$$

let $$a=\dfrac{x}{z},b=\dfrac{y}{z}$$
then $$a^2+b^2+6=4(a+b)$$
and we only find this minimum
$$\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}+\sqrt{a^2+b^2}$$
use Cauchy-Schwarz inequality,we have
$$\left(\dfrac{a^3}{b(a+1)^2}+\dfrac{b^3}{a(b+1)^2}\right)(ab(a+1)^2+ba(b+1)^2)\ge (a^2+b^2)^2$$
then we only find this follow minimum 
$$\dfrac{(a^2+b^2)}{ab[(a+1)^2+(b+1)^2]}+\sqrt{a^2+b^2}$$
since 
$$\dfrac{(a^2+b^2)}{ab[(a+1)^2+(b+1)^2]}\ge\dfrac{2}{a^2+b^2+2(a+b)+2} $$because use 
$a^2+b^2\ge 2ab$
so we ony find
$$\dfrac{2}{a^2+b^2+2(a+b)+2}+\sqrt{a^2+b^2}=\dfrac{4}{3a^2+3b^2+10}+\sqrt{a^2+b^2}$$
let $x=\sqrt{a^2+b^2}\in[\sqrt{2},3\sqrt{2}]$
so
$$f(x)=\dfrac{4}{3x^2+10}+x$$
then follow is easy to find minimum.
