For real numbers $\displaystyle 0\lt x\lt \frac{\pi}{2}$, prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$. (First Solution)

For real numbers $\displaystyle 0\lt x\lt \frac{\pi}{2}$, prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$.

MarkFL's solution:

If we observe that:

$\displaystyle g(x)=\sin^2(x)\tan(x)$

$\displaystyle h(x)=\cos^2(x)\cot(x)$

are complimentary functions, we can state the problem as:

Optimize the objective function:

$\displaystyle f(x,y)=\sin^2(x)\tan(x)+\sin^2(y)\tan(y)$

Subject to the constraints:

$\displaystyle x+y=\frac{\pi}{2}$

$\displaystyle 0\lt x,\,y \lt \frac{\pi}{2}$

And so...wait for it...can you guess where I'm going with this?...Yes! By cyclic symmetry, we know the critical point is:

$\displaystyle (x,y)=\left(\frac{\pi}{4},\frac{\pi}{4}\right)$

And we then find:

$\displaystyle f\left(\frac{\pi}{4},\frac{\pi}{4}\right)=1$

Checking another point on the constraint, such as:

$\displaystyle (x,y)=\left(\frac{\pi}{6},\frac{\pi}{3}\right)$

We find:

$\displaystyle f\left(\frac{\pi}{6},\frac{\pi}{3}\right)=\frac{5}{2\sqrt{3}}>1$

And so we know:

$\displaystyle f_{\min}=1$