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 :/

Note that $a\not=b$ implies $(a-b)(a^n-b^n)\gt0$.  Using this and the inductive hypothesis, we get
$$\begin{align}
(a+b)^{n+1}&=(a+b)^n(a+b)\cr
&\lt2^{n-1}(a^n+b^n)(a+b)\cr
&=2^{n-1}(a^{n+1}+ab^n+a^nb+b^{n+1})\cr
&=2^n(a^{n+1}+b^{n+1})-2^{n-1}(a^{n+1}-ab^n-a^nb+b^{n+1})\cr
&=2^n(a^{n+1}+b^{n+1})-2^{n-1}(a-b)(a^n-b^n)\cr
&\lt2^n(a^{n+1}+b^{n+1})
\end{align}$$
