# Tangent, Cotangent and Square Roots Inequality

### Solution

Denote $\sqrt{3\cot A\cot B}=z,\;$ $\sqrt{3\cot A\cot C}=y,\;$ $\sqrt{3\cot B\cot C}=x.\;$ Then $x,y,z\gt 0\;$ and $x^2+y^2+z^2=3.\;$ The required inequality becomes

$\displaystyle(x+y+z)\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\ge 9.$

By the Cauchy-Schwarz inequality,

$\displaystyle(xy+yz+zx)\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\ge (x+y+z)^2,$

which is same as

$\displaystyle (x+y+z)\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\ge \frac{(x+y+z)^3}{xy+yz+zx}.$

Thus, suffice it to show that $\displaystyle\frac{(x+y+z)^3}{xy+yz+zx}\ge 9.\;$ Let $x+y+z=3s.\;$ By the AM-QM inequality, $s\le 1,\;$ and clearly $xy+yz+zx=3\displaystyle\left(\frac{3s^2-1}{2}\right),\;$ implying $\displaystyle s\gt\frac{1}{\sqrt{3}}.$

We need to prove that $\displaystyle\frac{2s^3}{3s^2-1}\ge 1,\;$ but this is equivalent to $(1-s)^2(1+2s)\ge 0,\;$ which it true.

### Acknowledgment

The problem that is due to Hung Nguyen Viet has been communicated to me by Leo Giugiuc, along with the above solution.