1. ## inequality

Let $\displaystyle a, \ b>0$. Prove that $\displaystyle \left(1+\frac{a}{b}\right)^n+\left(1+\frac{b}{a}\r ight)^n\geq 2^{n+1}, \ \forall n\in\mathbf{Z}$

I have proved for $\displaystyle n\geq 0$, but I'm stuck for n<0.

2. Originally Posted by red_dog
Let $\displaystyle a, \ b>0$. Prove that $\displaystyle \left(1+\frac{a}{b}\right)^n+\left(1+\frac{b}{a}\r ight)^n\geq 2^{n+1}, \ \forall n\in\mathbf{Z}$

I have proved for $\displaystyle n\geq 0$, but I'm stuck for n<0.
Just out of curiosity, how did you prove the result for $\displaystyle n\geqslant0$. Did you use a concavity argument? If so, I think you can extend it to negative values.

3. Consider $\displaystyle f(x)=x^n$ and apply Jensen's inequality.

4. Originally Posted by Krizalid
Consider $\displaystyle f(x)=x^n$ and apply Jensen's inequality.
Isn't that like what I said about using the concavity?

5. Didn't see your post, sorry.

Actually, it's even easier, we can tackle this without Jensen.

We have $\displaystyle \left( 1+\frac{a}{b} \right)\left( 1+\frac{b}{a} \right)=2+\frac{a}{b}+\frac{b}{a}\ge 4,$ hence $\displaystyle \left( 1+\frac{a}{b} \right)^{n}+\left( 1+\frac{b}{a} \right)^{n}\ge 2\sqrt{4^{n}}=2^{n+1}.$