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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 !

  • 0
    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}$2012-03-27

5 Answers 5