How to show $(a + b)^n \leq a^n + b^n$, where $a, b \geq 0$ and $n \in (0, 1]$? Does anyone happen to know a nice way to show that $(a+b)^n \le a^n+b^n$, where $a,b\geq 0$ and $n \in (0,1]$?  I figured integrating might help, but I've been unable to pull my argument full circle. Any suggestions are appreciated :) 
 A: Let $f(x)=x^n$, where $n\in(0,1]$ and $a\geq b$.
Hence, $f$ is a concave function and $(a+b,0)\succ(a,b)$.
Thus, by Karamata $$(a+b)^n+0^n\leq a^n+b^n$$ 
and we are done!
A: We have $$ 1 = \frac{a}{a+b} +\frac{b}{a+b} \le \left(\frac{a}{a+b}\right)^n +\left(\frac{b}{a+b} \right)^n $$ for $0<n\le 1.$
A: The cases $a=0$ is trivial. For $a\ne 0$ let $m=1/n$ and $c=a^{1/m}=a^n$ and $d=b^{1/m}=b^n.$ Then $$(a+b)^n\leq a^n+b^n\iff (c^m+d^m)^{1/m}\leq c+d\iff$$ $$\iff  c^m+d^m\leq (c+d)^m\iff 1+(d/c)^m\leq (1+d/c)^m\iff$$ $$\iff (d/c)^m\leq (1+d/c)^m-1\iff$$ $$\iff  m\int_0^{d/c}x^{m-1}dx\leq m\int_0^{d/c}(1+x)^{m-1}dx$$ which holds as $d/c\geq 0$ and $m-1\geq 0.$
A: Assume that
$a \ge b$.
If $a = b = 0$,
the result is immediate.
If $a > 0$,
divide
$(a+b)^n \le a^n+b^n
$
by
$a^n$
to get
$(1+b/a)^n \le 1+(b/a)^n
$.
Since
$b \le a$,
$0 \le b/a \le 1$,
so this becomes
$(1+x)^n \le 1+x^n
$
where
$x = b/a$.
Let $f(x)
=1+x^n-(1+x)^n
$.
$f(0) = 0$
and
$f(1)
=2-2^n
\ge 0
$
since
$0 < n \le 1$.
$f'(x)
=nx^{n-1}-n(1+x)^{n-1}
=n(x^{n-1}-(1+x)^{n-1})
=n(\frac1{x^{1-n}}-\frac1{(1+x)^{1-n}})
$.
Since
$0 < n \le 1$,
$1-n \ge 0$
so that
$x^{1-n}
\le (1+x)^{1-n}
$
so that
$\frac1{x^{1-n}}\ge\frac1{(1+x)^{1-n}}
$
so that
$f'(x) \ge 0$.
Since
$f(0) = 0$
and
$f'(x) \ge 0$
for $0 < x\le 1$,
$f(x)
\ge 0
$
for
$0 \le x \le 1$.
Note that
the inequality goes the other way
if $n > 1$.
