How to show that $f(x) = \frac{\sqrt{\cos x}}{1-x^2}$ is convex and has a minimum value I want to show that  $f(x) = \frac{\sqrt{\cos x}}{1-x^2}$ is convex on the interval $]-1,1[$. How do I have to proceed?
I did take the derivative of the function which is 
$f'(x) = \frac{2x\sqrt{\cos x }}{(1-x^2)^2}-\frac{\sin x}{2\sqrt{\cos x}(1-x^2)}$
For a function to be convex I need to show that it is increasing on the interval?
I'm a little stuck to proof that rigorously. Thanks for your help!
 A: Let $f$ be given by 
$$f(x)=\frac{\sqrt{\cos x}}{1-x^2}$$
for $|x|<1$.  Then, the first derivative of $f$ is 
$$f'(x)=\frac{\sqrt{\cos x}}{2(1-x^2)}\left(\frac{4x}{1-x^2}-\tan x\right) \tag 1$$
From $(1)$ we can write
$$\frac{f'(x)}{f(x)}=\frac12\left(\frac{4x}{1-x^2}-\tan x\right) \tag 2$$
whereupon differentiating both sides of $(2)$ and solving for $f''(x)$ reveals that 
$$f''(x)=\frac{\left(f'(x)\right)^2}{f(x)}+f(x)\left(\frac{4(1+x^2)}{(1-x^2)^2}-\sec^2 x\right)$$
It is evident that $f(x)>0$ and $\left(f'(x)\right)^2\ge 0$ for $|x|<1$.  We will now show that the term $\left(\frac{4(1+x^2)}{(1-x^2)^2}-\sec^2 x\right)$ is positive thereby establishing that $f''(x)> 0$.
Recall that the sine function satisfies the inequality 
$$|\sin x|\le |x| \tag 2$$
From $(2)$ it is easy to show that for $|x|\le 1$
$$\cos x\ge \sqrt{1-x^2} \tag 3$$
Using $(3)$ reveals
$$\begin{align}
\left(\frac{4(1+x^2)}{(1-x^2)^2}-\sec^2 x\right) &\ge \left(\frac{4(1+x^2)}{(1-x^2)^2}-\frac{1}{1-x^2}\right)\\\\
&=\frac{3+5x^2}{(1-x^2)^2}\\\\
&>0
\end{align}$$
Since $f''(x)>0$, $f$ is convex on $-1<x<1$.  
Since $f$ is even and convex, it attains it minimum at $x=0$ where we have $f(0)=1$.
