Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n})$

The simpliest proof:

as $x\to x^n$ is convex, $$ \left(\frac {x+y}2\right)^n \le \frac {x^n+y^n}2 $$ Now multiply by $2^n$ and you are done.

First of all, this is only true for positive $a$, $b$.

For the induction step: suppose $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ for all $a,b>0$. Multiplying both sides of the inductive hypothesis by $a+b$ (here we use $a+b>0$), we find $(a+b)^{n+1} \leq 2^{n}(a^{n} + b^{n})(a+b)$. It remains to show that $(a^n+b^n)(a+b) \leq 2(a^{n+1}+b^{n+1})$. This is equivalent to $(a^n-b^n)(a-b) \geq 0$, which is true as $x \mapsto x^n$ is an increasing function.

This is the generalized mean inequality for two variables (comparing the arithmetic mean and the $n$-th power mean).

$$(a+b)^n\leq 2^{n-1}(a^n+b^n)\Longleftrightarrow\left(\frac{a+b}2\right)^n\leq\frac{a^n+b^n}2\Longleftrightarrow\frac{a+b}2\leq\sqrt[n]\frac{a^n+b^n}2$$ Which is a special case of the generalized mean inequality if $a,b\geq0$ and $n\geq1$ (can even be real).

The full inequality gives you a nice generalized bound $$\left(\sum_{i=1}^k a_i\right)^n\leq k^{n-1}\sum_{i=1}^k a_i^n$$