# Inequality with condition $x^2+y^2+z^2=1$.

let $x,y,z>0$ such that $x^2+y^2+z^2=1$. Find the minimum of $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$

Is the answer $3\sqrt{3}$ by any chance?

#### Solutions Collecting From Web of "Inequality with condition $x^2+y^2+z^2=1$."

Apply AM-GM and CS inequalities:$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \geq \dfrac{9}{x+y+z} \geq \dfrac{9}{\sqrt{3}\cdot (x^2+y^2+z^2)}= 3\sqrt{3}$$

Just another way, sum three AM-GMs of form:
$$\frac1x + \frac1x +3\sqrt{3}x^2 \ge 3\sqrt3$$
Equality is iff $x = \frac1{\sqrt3} = y= z$