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 $f(x,y,z)=\dfrac{xz}{y}+\dfrac{yz}{x}+\dfrac{xy}{z}$.   

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.
