Conditional extreme value of a function Let $x,y,z$ be the positive real numbers, if $x^2+y^2+z^2=1$, then how can we find the minimal value of this function $f(x,y,z)=\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}$.   
 A: I will show that, 
$$ \frac{xy}{z}+ \frac{yz}{x}+\frac{xz}{y} \ge \sqrt{3}$$ 
This is equivalent to, 
$$ (\frac{xy}{z}+ \frac{yz}{x}+\frac{xz}{y})^2 \ge 3$$ 
Which is same as,
$$ \frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2} + 2(x^2+y^2+z^2) \ge 3 = 3(x^2+y^2+z^2) $$ 
So it suffice to prove that, 
$$ \frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2} \ge x^2+y^2+z^2 $$ 
This is trivial if someone notices inequality, $a^2+b^2+c^2 \ge ab+bc+ac $
Elaborating, 
Note that, by AM-GM,Following inequalities are true, 
$$\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2} \ge 2y^2 $$
$$\frac{y^2z^2}{x^2}+\frac{x^2z^2}{y^2} \ge 2z^2 $$ 
$$\frac{x^2y^2}{z^2}+\frac{x^2z^2}{y^2} \ge 2x^2 $$ 
Adding gives the desired result. 
Equality when, $x=y=z \implies x=y=z=\frac{1}{\sqrt{3}}$
A: Hint First, note that $$\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}=xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).$$ Using $a+b+c\geq 3\sqrt[3]{abc}$, try to estimate lower bounds for $xyz$ and $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$. If you are stuck in the second, consider $$(x^2+y^2+z^2)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).$$ Show that the conditions for minimizing $xyz$ and $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$ can happen at the same time.
