Minimize $P=5\left(x^2+y^2\right)+2z^2$ For $\left(x+y\right)\left(x+z\right)\left(y+z\right)=144$, minimize $$P=5\left(x^2+y^2\right)+2z^2$$
I have no idea. Can you make a few suggestions? 
 A: HINT: prove that $$P=5(x^2+y^2)+2z^2\geq 72$$ and the equal sign will atained at $$(x,y,z)=(2,2,4)$$
A: By AM-GM 
$1728=(3x+3y)(2x+2z)(2y+2z)\leq \left(\frac{5x+5y+4z}{3}\right)^3$
$\Rightarrow 5x+5y+4z\geq 36$
By Cauchy-Schwarz we have:
$18P=(5x^2+5y^2+2z^2)(5+5+8)\geq (5x+5y+4z)^2\geq 36^2$
$$\Rightarrow P\geq 72$$
Hence $P_{\min}=72\Leftrightarrow (x,y,z)=(2,2,4)$
A: Let $x=y=2$ and $z=4$.
Hence, $P=72$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$5(x^2+y^2)+2z^2\geq72\left(\sqrt[3]{\frac{(x+y)(x+z)(y+z)}{144}}\right)^2.$$
It's enough to prove last inequality for non-negative variables.
Let $x+y=tz$.
Since $x^2+y^2\geq\frac{1}{2}(x+y)^2$ and $(x+z)(y+z)\leq\frac{1}{2}(x+y+2z)^2$, it remains to prove that
$$\frac{5}{2}t^2+2\geq72\left(\sqrt[3]{\frac{\frac{t(t+2)^2}{4}}{144}}\right)^2$$ or
$$(5t^2+4)^3\geq9t^2(t+2)^4,$$
which is C-S and AM-GM:
$$(5t^2+4)^3=\left(\frac{(5+4)(5t^2+4)}{9}\right)^3\geq\left(\frac{(5t+4)^2}{9}\right)^3=$$
$$=\left(\frac{(3t+2(t+2))^2}{9}\right)^3\geq\left(\frac{\left(3\sqrt[3]{3t\cdot(t+2)^2}\right)^2}{9}\right)^3=9t^2(t+2)^4.$$
Done!
