Prove that, for any positive integer n: $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n})$ Prove that, for any positive integer n:  $(a + b)^{n} \leq 2^{n-1}(a^{n}+b^{n}) $
I tried induction theorem,
when $n = 1$ it is obviously right. But, 
say $n=k$, 
It does not make sense since I cannot expand the $2^{k-1}$($a^{k}$+$b^{k}$).
And I also looked through this similar question but no help. 
Prove: $(a + b)^{n} \geq a^{n} + b^{n}$
 A: 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: $$(a+b)^n\leq 2^{n-1}(a^n+b^n)\Longleftrightarrow\left(\frac{a+b}2\right)^n\leq\frac{a^n+b^n}2\Longleftrightarrow\frac{a+b}2\leq\sqrt[n]\frac{a^n+b^n}2$$
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
$$\left(\sum_{i=1}^k a_i\right)^n\leq k^{n-1}\sum_{i=1}^k a_i^n$$
A: 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.
A: Also, the TL method helps.
Indeed, since our inequality is homogeneous, we can assume that $a+b=2$ and we need to prove that
$$a^n+b^n\geq2$$ or
$$(a^n-na+n-1)+(b^n-nb+n-1)\geq0,$$ which is true by AM-GM:
$$\sum_{cyc}(a^n-nb+n-1)\geq\sum_{cyc}\left(n\sqrt[n]{a^n\cdot1^{n-1}}-na\right)=0$$ and we are done!
