How to prove this inequality $2\cos{x} + \sec^{2}{x} - 3 \ge{0}$ How to prove this inequality in interval: $(0,\frac{\pi}{2})$
$
2\cos{x} + \sec^{2}{x} - 3 \ge{0}
$
I have to use it to show that the differentiation of a function is greater than or equal to 0 so that I can prove that the function is an increasing function.
Edit: Added the interval
 A: As pointed out by @Andrew Chin that it does not hold for all real numbers. Thus it is reasonable to restrict the domain to some meaningful interval such as $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ so that $\cos x > 0$. In that case, using the AM-GM inequality you have: $LHS = \cos x+ \cos x + \sec^2x - 3 \ge 3\sqrt[3]{\cos x\cdot \cos x\cdot \sec^2 x} - 3 = 3 -3 = 0 = RHS $.
A: By Fermat's theorem, we examine zeroes of derivative of the function $f(x)=2\cos{x} + \sec^{2}{x} - 3 $ to find the extrema, that is, to solve the following:
$$
f'(x)=2 \tan x \sec ^2 x-2 \sin x = 0
$$
The solution is $x=\{k\pi | k\in\mathbb Z\}$. When $x=\pi$, we get
$$
2 \cos (x)+\sec ^2(x)-3=-4
$$
Thus $2\cos{x} + \sec^{2}{x} - 3 \ge{0}$ doesn't hold.

Since the interval is added, we can prove the claim is true. For $x=0$, $f(x)=0$. And when $x\in(0,\pi/2)$, it is trivial that $f'(x)\geq 0$, thus $f(x)$ monotonously increases on $(0,\pi/2)$. Hence $f(x)\geq f(0)=0$ for all $x\in(0,\pi/2)$ and we finished the proof.
A: We have
$$2 \cos{x} + \sec^{2}{x} - 3 = 2 (\cos x - 1) + \left(\frac{1}{\cos^2 x} - 1\right) = \frac{(2 \cos x+1)(\cos x-1)^2}{\cos^2x}.$$
Thefore the inequality does not hold.
