Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either 
$(1) 0 \leq a,b \leq 1$ 
OR 
$(2) ab \geq 3$
Since this question was under Trigonometry, I assumed the following.
Since $a,b$ are positive real numbers with $0 \leq a,b \leq 1$, I can assume that for some $x,y, a=\tan(x), b=\tan(y)$ and therefore it is to be shown that 
$$\frac{1}{\sec x} + \frac{1}{\sec y} = \cos x+ \cos y \geq \frac{2\cos x \cos y}{\sqrt{cos(x-y)}}$$ 
(Originally posted without that $2$ on the right - Sorry!)
I do know that 
$$\cos x + \cos y \geq 2 \sqrt{\cos x \cos y}$$ 
Now how to proceed? Just give me hints !

The statement seems wrong: Put $a:=0.1$, $b:=1$. Then the left side is
$${1\over\sqrt{1.01}}+{1\over\sqrt{2}}\doteq1.702\ ,$$
and the right side is
$${2\over\sqrt{1.1}}\doteq 1.907\ .$$
