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 !
 A: Edit: Note that when the answer below was posted, and for a long time thereafter, the expression on the right was $\dfrac{1}{\sqrt{1+ab}}$.
I am puzzled. Suppose without loss of generality that $a\le b$. Then $1+a^2\le 1+ab$, and therefore $\dots$.  So the desired inequality is true for any positive $a$, $b$.
A: 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
$${2\over\sqrt{1.1}}\doteq 1.907\ .$$
A: The inequality holds with the inequality sign reversed in the first case.
Indeed suppose that $a \in [0,1]$ and define $f \colon [0,1] \to \mathbb{R}$ by
$$f(b) = \frac{2}{\sqrt{1 + ab}} - \frac{1}{\sqrt{1 + a^2}} - \frac{1}{\sqrt{1 + b^2}},$$
so that the claim amounts to showing that $f(b) \ge 0$ for all $b \in [0,1]$. This can be done in a standard way by looking at the zeros of the derivative.
In the second case ($ab \ge 3$) it seems to me that the inequality sign is correct as stated, although I didn't prove it. (I just plotted it with respect to $a$ after setting $ab = 3$.) The proof (if the claim is true) should go with a similar argument as in the first part, however.
A: 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. 
A: Let $f(x)=\frac{1}{\sqrt{1+e^{2x}}}$.
Hence, $f''(x)=\frac{e^{2x}(e^{2x}-2)}{\sqrt{(e^{2x}+1)^5}}\geq0$ for all $x\geq\frac{1}{2}\ln2$.
Id est, for all $\{a,b\}\subset[\sqrt2,+\infty)$ by Jensen we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{1+a^2}}=\sum_{cyc}f(\ln{a})\geq2f\left(\frac{\ln{a}+\ln{b}}{2}\right)=\frac{2}{\sqrt{1+ab}}$$
