Remove the $\tan x$ and $\cot x$ by replacing them with $\frac{\sin x}{\cos x}$ and $\frac{\cos x}{\sin x}$ respectively. Then it is equivalent to proving $\frac{\cos^3x}{\sin x}+\frac{\sin^3x}{\cos x}\ge 1$. But $\frac{\cos^3x}{\sin x}+\frac{\sin^3x}{\cos x}\ge\cos^2 x+\sin^2 x=1$ by Muirhead ($\cos x$ and $\sin x$ are both positive), so we are done.

WakeUp wrote: Let $x$ be a real number such that $0<x<\frac{\pi}{2}$. Prove that \[\cos^2(x)\cot (x)+\sin^2(x)\tan (x)\ge 1\] This is equivalent with $1-2\sin^{2}x\cdot \cos^{2}x \geq \sin x\cos x$ $-2\sin^{2}x\cdot \cos^{2}x-\sin x\cos x+1 \geq 0$ $-1\leq \sin x\cos x \leq \frac{1}{2}$ From AM-GM we have $\sin x \cos x\leq \frac{1}{2}$ from this we can see that the inequality is true.

$ \frac{\cos^{3}x}{\sin x}+\frac{\sin^{3}x}{\cos x}\ge 1 $ can be also done be Rearrangement.

${ \frac{\cos^{3}x}{\sin x}+\frac{\cos^{3}x}{\sin x}+sin^{2}x}\ge 3cos^{2}x $ ${ \frac{\sin^{3}x}{\cos x}+\frac{\sin^{3}x}{\cos x}+cos^{2}x}\ge 3sin^{2}x $

by Carlson $(\frac{cos^3x}{sinx}+\frac{sin^3(x)}{cosx})(sinxcosx+cosxsinx)\ge (sin^2x+cos^2x)^2=1$ trivial

$cos^2x cotx+sin^2x tanx=\frac{cos^3x}{sinx}+sinxcosx+\frac{sin^3x}{cosx}+cosxsinx-sin2x$ $\ge 2cos^2x+2sin^2x-sin2x=2-sin2x\ge 1.$ \[\Rightarrow cos^2x cotx+sin^2x tanx\ge 1.\] Similar to Baltic Way 2010 Generalization of Baltic Way 2010

$cos^2x cotx+sin^2x tanx=\frac{cos^4x}{sinxcosx}+\frac{sin^4x}{sinxcosx}$ $\ge \frac{(cos^2x+sin^2x)^2}{2sinxcosx}\ge \frac{1}{sin^2x+cos^2x}=1.$ \[\Rightarrow cos^2x cotx+sin^2x tanx\ge 1.\] Similar to Baltic Way 2010 Generalization of Baltic Way 2010