A very tricky trig problem from China !! RMO or IITJEE Advanced mathematics training

Problem: (China 2003)

Let n be a fixed positive integer. Determine the smallest positive real number \lambda such  that for any \theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n} in the interval (0,\frac{\pi}{2}), if

\tan {\theta_{1}} \tan {\theta_{2}}\ldots \tan_{\theta_{n}}=2^{\frac{n}{2}}, then

\cos {\theta_{1}}+\cos{\theta_{2}}+\ldots+\cos{\theta_{n}} \leq \lambda.

Solution:

(by Yumin Huang)

The answer is:

\lambda = \frac{\sqrt{3}}{3}, if n=1

\lambda = \frac{\sqrt{2\sqrt{3}}}{3}, if n=2

\lambda = n-1, if n \geq 3

The case n=1 is trivial.

If n=2,we claim that \cos{\theta_{1}}+\cos{\theta_{2}} \leq \frac{2\sqrt{3}}{3} with  equality if and only if \theta_{1}=\theta_{2}=\arctan{\sqrt{2}}. It suffices to show that

\cos^{2}{\theta_{1}}+\cos^{2}{\theta_{2}}+2\cos{\theta_{1}}\cos{\theta_{2}} \leq \frac{4}{3}

or,

\frac{1}{1+\tan^{2}{\theta_{1}}}+\frac{1}{1+\tan^{2}{\theta_{2}}}+ 2 \sqrt{\frac{1}{(1+\tan^{2}{\theta_{1}})(1+\tan^{2}{\theta_{2}})}} \leq \frac{4}{3}

Because \tan{\theta_{1}}\tan{\theta_{2}}=2, we get

(1+\tan^{2}{\theta_{1}})(1+\tan^{2}{\theta_{2}})=5+\tan^{2}{\theta_{1}}+\tan^{2}{\theta_{2}}.

By setting,

\tan^{2}{\theta_{1}}+\tan^{2}{\theta_{2}}=x, the last inequality becomes

\frac{2+x}{5+x}+2\sqrt{\frac{1}{5+x}} \leq \frac{4}{3}, or

2\sqrt{\frac{1}{5+x}} \leq \frac{14+x}{3(5+x)}

Squaring both sides and clearing denominators, we get 36(5+x) \leq 196+ 28x +x^{2}, that is,

0 \leq x^{2}-8x+16=(x-4)^{2}. This  establishes our claim.

Now, assume that n \geq 3. We claim that \lambda=n-1. Note that \lambda \geq n-1; by setting \theta_{2}=\theta_{3}= \ldots = \theta_{n}=\theta and letting \theta \rightarrow 0, we find that \theta_{1} \rightarrow \frac{\pi}{2}, and so the left hand side of the desired inequality approaches n-1. It suffices to show  that

\cos{\theta_{1}}+\cos{\theta_{2}}+\ldots+\cos{\theta_{n}} \leq n-1.

Without loss of generality, assume that \theta_{1} \geq \theta_{2} \geq \ldots \theta_{n}. Then,

\tan{\theta_{1}}\tan{\theta_{2}}\tan{\theta_{3}} \geq 2\sqrt{2}

It suffices to show that \cos{\theta_{1}}+\cos{\theta_{2}}+\cos{\theta_{3}} <2 relation *

But, because \sqrt{1-x^{2}} \leq 1 - \frac{1}{2}x^{2}, \cos{\theta_{i}}=\sqrt{1-\sin^{2}{\theta_{i}}} < 1-\frac{1}{2}\sin^{2}{\theta_{i}}

Consequently, by the arithmetic geometric mean inequality, 

\cos{\theta_{2}}+\cos{\theta_{3}} < 2-\frac{1}{2}(\sin^{2}{\theta_{2}}+\sin^{2}{\theta_{3}}) \leq 2 - \sin{\theta_{2}}\sin{\theta_{3}}

Because \tan^{2}{\theta_{1}} \geq \frac{8}{\tan^{2}{\theta_{2}}\tan^{2}{\theta_{3}}}

we have \sec^{2}{\theta_{1}} \geq \frac{8+\tan^{2}{\theta_{2}}\tan^{2}{\theta_{3}}}{\tan^{2}{\theta_{2}}\tan^{2}{\theta_{3}}}

or $latex $

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