Too Easy (or too Difficult?)

Geometry Level 4

Determine the minimum value of

$\frac{\sec^4\alpha}{\tan^2\beta}+\frac{\sec^4\beta}{\tan^2\alpha}$

over all $\alpha,\beta \neq \frac{k\pi}{2}$ where $k\in\mathbb{Z}$ .

Details and Assumptions:

$\mathbb{Z}$ is the set of integers.

The answer is 8.

1 solution

Ameya Salankar
Jun 17, 2014

Set $a=\tan^2\alpha$ and $b=\tan^2\beta$ . The problem now reduces to finding the minimum value of

$\frac{(a+1)^2}{b}+\frac{(b+1)^2}{a}$ , with $a,b \geq 0$ .

We have,

$\frac{(a+1)^2}{b}+\frac{(b+1)^2}{a} = \frac{a^2+2a+1}{b}+\frac{b^2+2b+1}{a}$ $= \left(\frac{a^2}{b}+\frac{1}{b}+\frac{b^2}{a}+\frac{1}{a}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)$

Applying the A.M.-G.M. inequality, we get-

$\left(\frac{a^2}{b}+\frac{1}{b}+\frac{b^2}{a}+\frac{1}{a}\right)+2\left(\frac{a}{b}+\frac{b}{a}\right)\geq4\sqrt[4]{\frac{a^2}{b}\times\frac{1}{b}\times\frac{b^2}{a}\times\frac{1}{a}}+4\sqrt{\frac{a}{b}\times\frac{b}{a}} = \boxed{8}$ .

You could apply AM-GM directly to each term, to get

$\frac{ (a+1)^2}{b} + \frac{ (b+1)^2}{a} \geq \frac{ 4a}{b} + \frac{4b}{a} \geq 8$

Equality holds if and only if $a = b = 1$ .

Calvin Lin Staff - 6 years, 12 months ago

Hmm... yeah! Wonder why I didn't think of that earlier.

Ameya Salankar - 6 years, 11 months ago

Too Easy (or too Difficult?)

Details and Assumptions:

The answer is 8.

1 solution

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