## 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) $\lambda = \frac{\sqrt{3}}{3}$, 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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