Binomial expansion lower bound $A^n + B^n \le (A+B)^n$ for non-integer $n$

Question

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}

Answer 1

Alt. hint: $\;\displaystyle \left(\frac{A}{A+B}\right)^n + \left(\frac{B}{A+B}\right)^n \le \frac{A}{A+B}+\frac{B}{A+B}=1\,$ when $\,n \ge 1\,$.

Answer 2

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 $x > 0$.