<s>[INST] Binomial expansion lower bound $A^n + B^n \le (A+B)^n$ for non-integer $n$ By the Binomial expansion for integer powers,
$$ (A+B)^n = \sum_{k=0}^n {n\choose{k}} A^{n-k} B^{k}$$
(I'm assuming $A,B\ge 0$) and so we get the easy estimate $A^n + B^n \le (A+B)^n$ for any positive integer $n$. 
Now what happens when we take $n$ to be non-integer? Well if $0<n\le1$, then we have the reverse inequality, 
 $$(A+B)^n \le A^n + B^n ,\qquad {(0<n\le 1)} $$
What about if $n>1$ is a non-integer? Can we still say $A^n + B^n \le (A+B)^n$? I'm pretty sure the answer is yes, but I wanted to check with you guys. Here is my quick argument.  Let $n = p + \alpha$ where $p \in \mathbb{N}$ and $0<\alpha<1$. 
\begin{align} A^n+B^n &= A^\alpha A^p + B^\alpha B^p \\ &\le \text{max}(A^\alpha,B^\alpha) (A^p+B^p) \\ &\le \text{max}(A^\alpha,B^\alpha) (A+B)^p \\ &\le (A+B)^\alpha (A+B)^p \\ &= (A+B)^n\end{align}
 [/INST] If $A=B$,
$A^n + B^n \le (A+B)^n$
is
$2A^n \le (2A)^n
=2^nA^n$
or
$2 \le 2^n$\which is true since
$n \gt 1$.
Assuming
$A \lt B$,
dividing by
$B^n$
and letting
$A/B = x$,
$A^n + B^n \le (A+B)^n$
becomes
$1+x^n \le (1+x)^n$
with
$0 < x < 1$.
Let
$f(x)
= (1+x)^n-1-x^n$.
$f(0) = 0$.
$f'(x)
=n(1+x)^{n-1}-nx^{n-1}
=n((1+x)^{n-1}-x^{n-1})
\gt 0$
since
$n-1 > 0$
so
$(1+x)^{n-1}
\gt x^{n-1}$.
Therefore
$f(x) > 0$
for $x > 0$.
</s>