Proving an inequality for all natural numbers involving parameters

Question

The problem asks to prove the following inequality:

$$\forall a,b>0,a\ne b\;\forall n\in \Bbb N, n>1:2^{n-1}(a^n+b^n)>(a+b)^n$$

I'd appreciate a hint, because I tried induction but it lead nowhere, so there has to be some other way to prove this or maybe even by induction but I believe it won't be straightforward.

Answer 1

Using $a,b> 0$, we have $$\tag1a^n+b^n=a^ka^{n-k}+b^kb^{n-k}\ge a^kb^{n-k}+a^{n-k}b^k$$ with equality iff $k=0$ or $k=n$ (because $a\ne b$). If you multiply out $(a+b)^n$, you get $2^n$ summands (one for each subset of $\{1,\ldots,n\}$, which can be grouped into $2^{n-1}$ pairs like $a^kb^{n-k}+a^{n-k}b^k$ (with the corresponding substs being complements of each other). For some subsets we have e.g. $k=1$, making $(1)$ strict for these cases. Therefor $$ 2^{n-1}(a^n+b^n)> (a+b)^n$$

Answer 2

Hint: Check Power Mean Inequality and use in the form: $$\sqrt[n]{\frac{a^n+b^n}{2}} > \frac{a+b}{2}$$

Answer 3

Alternately by Holder's inequality: $$(a^n+b^n) \underbrace{(1+1)(1+1)...(1+1)}_{n-1 \;\text{times}} > (a+b)^n$$