Proof: $2^{n-1}(a^n+b^n)>(a+b)^n$ If $n \in \mathbb{N}$ with $n \geq 2$ and $a,b \in \mathbb{R}$ with $a+b >0$ and $a \neq b$, then $$2^{n-1}(a^n+b^n)>(a+b)^n.$$
I tried to do it with induction. The induction basis was no problem but I got stuck in the induction step: $n \to n+1$
$2^n(a^{n+1}+b^{n+1})>(a+b)^{n+1} $
$ \Leftrightarrow 2^n(a\cdot a^n + b\cdot b^n)>(a+b)(a+b)^n$
$\Leftrightarrow a(2a)^n+ b(2b)^n>(a+b)(a+b)^n$
dont know what to do now :/

Start from the other side:
$$(a+b)^{n+1} < (a+b)(2^{n-1}(a^n+b^n))= 2^{n-1}a^{n+1}+2^{n-1}a^{n}b+2^{n-1}ab^{n}+2^{n-1}b^{n+1}$$
Now, prove that $(a^{n}-b^{n})(a-b)>0$ 
which implies
$$2^{n-1}a^{n}b+2^{n-1}ab^{n}< 2^{n-1}a^{n+1}+2^{n-1}b^{n+1} \,.$$
P.S. If you are familiar, the last inequality:
$$a^{n}b+ab^{n}< a^{n+1}+b^{n+1} $$
also follows immediately from the AM-GM inequality.
