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 !

• i guess you may solve that quickly if you use complex numbers: $x=1+ia = u*e^{i\alpha}$ and $y=1+ib = v*e^{i\beta}$ – Hassan Mar 27 '12 at 2:38

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 $a$, $b$.

• +1! I was about to comment, it seems the OP might have the problem wrong... – Aryabhata Mar 27 '12 at 3:47
• This question is from a book by Titu Andreescu and Zuming Feng (Someone told me). After searching on google I found the solution on scribd.com (I think the right side should have a 2). – Kirthi Raman Mar 27 '12 at 11:20

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 $${2\over\sqrt{1.1}}\doteq 1.907\ .$$

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.

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.

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: $$\sum_{cyc}\frac{1}{\sqrt{1+a^2}}=\sum_{cyc}f(\ln{a})\geq2f\left(\frac{\ln{a}+\ln{b}}{2}\right)=\frac{2}{\sqrt{1+ab}}$$