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 !

Nice approach, you are very very close but this is how you finish it off: let $u^2=\cos x$ and $t^2=\cos y$ . Substituting, you have $u^2+t^2 \ge 2ut$, or  $u^2-2ut+t^2 \ge 0$ , and this can be factored into $(u-t)^2 \ge 0$ , which is always true of course. 
