1. ## Prove

Prove :
If :$\displaystyle \left\{ \begin{array}{l} x,y \in \Re+ \\ n \in {\rm N} \\ x + y = 1 \\ \end{array} \right.$
Then :$\displaystyle \left( {1 + \frac{1}{{x^n }}} \right)\left( {1 + \frac{1}{{y^n }}} \right) \ge \left( {1 + 2^n } \right)^2$

2. Hint: set $\displaystyle x= \cos^2 \theta, y = \sin^2 \theta$. Then

$\displaystyle \left(1+\frac{1}{x^n}\right)\left(1+\frac{1}{y^n}\ right)$ is minimized when $\displaystyle \theta = \pi/4$ so that $\displaystyle x=y=\cos^2\theta = 1/2$...

3. Originally Posted by Bruno J.

Hint: set $\displaystyle x= \cos^2 \theta, y = \sin^2 \theta$. Then

$\displaystyle \left(1+\frac{1}{x^n}\right)\left(1+\frac{1}{y^n}\ right)$ is minimized when $\displaystyle \theta = \pi/4$ so that $\displaystyle x=y=\cos^2\theta = 1/2$...

Hello : Thank you, Are you the details

4. Hehehe, I try to help where I can

That hint is more helpful ... did you use calculus to find the minimum?

Cuz you could also use trig identities to get:

$\displaystyle \left(1+\frac{1}{x^n}\right)\left(1+\frac{1}{y^n}\ right) = 1+ \displaystyle{\frac{2^{2n}}{\sin^{2n}2 \phi}} + \displaystyle{\frac{2^{2n}}{\sin^{2n}2 \phi}} \geq 1 + 2^{n+1} + 2^{2n} = (1+2^n)^2$