# How to prove this inequality in interval: (0,\frac{\pi}{2}) \ \

How to prove this inequality in interval: $$\displaystyle{\left({0},{\frac{{\pi}}{{{2}}}}\right)}\ \ {2}{\cos{{x}}}+{{\sec}^{{2}}{x}}-{3}\geq{0}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Thomas Nickerson
Restrict the domain to some meaningful interval such as $$\displaystyle{x}\in{\left(-{\frac{{\pi}}{{{2}}}},{\frac{{\pi}}{{{2}}}}\right)}$$ so that $$\displaystyle{\cos{{x}}}{>}{0}$$. In that case, using the AM-GM inequality you have:
$$\displaystyle{L}{H}{S}={\cos{{x}}}+{\cos{{x}}}+{{\sec}^{{2}}{x}}-{3}\geq{3}{\sqrt[{{3}}]{{{\cos{{x}}}\cdot{\cos{{x}}}\cdot{{\sec}^{{2}}{x}}}}}-{3}={3}-{3}={0}={R}{H}{S}$$
###### Not exactly what you’re looking for?
Papilys3q
We have
$$\displaystyle{2}{\cos{{x}}}+{{\sec}^{{2}}{x}}-{3}={2}{\left({\cos{{x}}}-{1}\right)}+{\left({\frac{{{1}}}{{{{\cos}^{{2}}{x}}}}}-{1}\right)}={\frac{{{\left({2}{\cos{{x}}}+{1}\right)}{\left({\cos{{x}}}-{1}\right)}^{{2}}}}{{{{\cos}^{{2}}{x}}}}}$$
Thefore the inequality does not hold.
Vasquez

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 $$2\cos x+\sec^2 x-3 \geq 0$$ doesn't hold.
Since the interval is added, we can prove the claim is true. For x=0, f(x)=0. And when $$x \in (0,\frac{\pi}{2})$$, it is trivial that $$f′(x) \geq 0$$, thus f(x) monotonously increases on $$(0,\frac{\pi}{2})$$. Hence $$f(x) \geq f(0)=0 \text{ for all } x\in (0,\frac{\pi}{2})$$ and we finished the proof.