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

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

MarkFL's solution:

If we observe that:

[MATH]g(x)=\sin^2(x)\tan(x)[/MATH]

[MATH]h(x)=\cos^2(x)\cot(x)[/MATH]

are complimentary functions, we can state the problem as:

Optimize the objective function:

[MATH]f(x,y)=\sin^2(x)\tan(x)+\sin^2(y)\tan(y)[/MATH]

Subject to the constraints:

[MATH]x+y=\frac{\pi}{2}[/MATH]

[MATH]0\lt x,\,y \lt \frac{\pi}{2}[/MATH]

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

[MATH](x,y)=\left(\frac{\pi}{4},\frac{\pi}{4}\right)[/MATH]

And we then find:

[MATH]f\left(\frac{\pi}{4},\frac{\pi}{4}\right)=1[/MATH]

Checking another point on the constraint, such as:

[MATH](x,y)=\left(\frac{\pi}{6},\frac{\pi}{3}\right)[/MATH]

We find:

[MATH]f\left(\frac{\pi}{6},\frac{\pi}{3}\right)=\frac{5}{2\sqrt{3}}>1[/MATH]

And so we know:

[MATH]f_{\min}=1[/MATH]