Inequality: if $\cos^2a+\cos^2b+\cos^2c=1$, prove $\tan a+\tan b+\tan c\geq 2(\cot a+\cot b+\cot c)$ Let $a,b,c$ be in $\left(0;\dfrac{\pi}2\right)$ such that $\cos^2a+\cos^2b+\cos^2c=1$. I am trying to prove the following inequality: $\tan a+\tan b+\tan c\geq 2\left(\cot a+\cot b+\cot c\right)$, but I do not know how. Does any know help me to show this?
 A: Lemma:
let $x,y,z>0$,then
$$\dfrac{3}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}\le\dfrac{x+y+z}{3}\le\sqrt{\dfrac{x^2+y^2+z^2}{3}}$$
let $$\cos{a}=x,\cos{b}=y,\cos{c}=z,x,y,z>0$$
Note
$$\tan{a}=\dfrac{\sin{a}}{\cos{a}}=\dfrac{\sqrt{1-\cos^2{a}}}{x}=\dfrac{\sqrt{y^2+z^2}}{x}\ge\dfrac{1}{\sqrt{2}}\dfrac{y+z}{x}$$
so
$$\tan{a}+\tan{b}+\tan{c}\ge\dfrac{1}{\sqrt{2}}\left(\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{y}{z}+\dfrac{z}{y}\right)$$
on the other hand,
$$\cot{a}=\dfrac{\cos{a}}{\sin{a}}=\dfrac{x}{\sqrt{y^2+z^2}}\le\dfrac{x\left(\dfrac{1}{y}+\dfrac{1}{c}\right)}{2\sqrt{2}}$$
so
$$\cot{a}+\cot{b}+\cot{c}\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{z}{x}+\dfrac{y}{z}+\dfrac{z}{y}\right)$$
so
$$\tan{a}+\tan{b}+\tan{c}\ge 2(\cot{a}+\cot{b}+\cot{c})$$
A: Since $a,b,c\in\left(0,\dfrac{\pi}{2}\right)$, and \[\cos^2a+\cos^2b+\cos^2c=1.\]We have\[a+b>\dfrac{\pi}{2},b+c>\dfrac{\pi}{2},c+a>\dfrac{\pi}{2}.\]Without loss of generality, we can assume that $a\geqslant b\geqslant c$, then\[1-3\cos^2a\geqslant1-3\cos^2b\geqslant 1-3\cos^2c,\]and\[\sin2a\leqslant \sin2b\leqslant \sin 2c\iff\frac{1}{\sin2a}\geqslant\frac{1}{\sin2b}\geqslant\frac{1}{\sin2c}.\]By Chebyshev's inequality, we have\begin{align}&\frac{1-3\cos^2a}{\sin2a}+\frac{1-3\cos^2b}{\sin2b}+\frac{1-3\cos^2c}{\sin2c}\\\geqslant&\frac{1}{3}\left(\frac{1}{\sin2a}+\frac{1}{\sin2b}+\frac{1}{\sin2c}\right)\left((1-3\cos^2a)+(1-3\cos^2b)+(1-3\cos^2c)\right)=0\end{align}Note that\begin{align}&\tan{a}+\tan{b}+\tan{c}- 2(\cot{a}+\cot{b}+\cot{c})\\=&2\left(\frac{1-3\cos^2a}{\sin2a}+\frac{1-3\cos^2b}{\sin2b}+\frac{1-3\cos^2c}{\sin2c}\right).\end{align}
