Proof for $a^c + b^c > (a + b)^c$ when $0 < c < 1$ and $a, b> 0$ and $1/c$ is nonintegral

Question

What is the proof for the statement $a^c + b^c > (a + b)^c$ when $0 < c < 1$, $a, b> 0$ and $1/c$ is non-integral? I have a very simple proof for this statement when $1/c$ is an integer (namely, just raise both sides to $1/c$ and the proof immediately follows from binomial expansion of $(a^c + b^c)^{1/c}$).

But what about the case when $1/c$ is non-integral?

This is just a special case of $|\cdot|{\ell_q}<|\cdot|{\ell_p}$ for $0<p<q<\infty$. See here: https://math.stackexchange.com/questions/4094/how-do-you-show-that-l-p-subset-l-q-for-p-leq-q/4122#4122 — Ben W, Dec 30 '18 at 04:56

score 3 · Accepted Answer · answered Dec 30 '18 at 05:21

3

$(1+t)^{c}-1-t^{c}$ vanishes at $t=0$ and is decreasing in $t>0$ (because its derivative is negative). Hence $(1+t)^{c}<1+t^{c}$. Put $t=\frac b a$ and muliply both sides by $a^{c}$.

answered Dec 30 '18 at 05:21

Kavi Rama Murthy

311,013

score 0 · Answer 2 · answered Jan 01 '19 at 19:55

0

The function $f:(0,\infty)\to \mathbb{R}$ defined by $f(x)=x^c$ with $0<c<1$ is strictly concave. Therefore $$f(a)+f(b)> f(a+b).$$

answered Jan 01 '19 at 19:55

user376343

8,311

Proof for $a^c + b^c > (a + b)^c$ when $0 < c < 1$ and $a, b> 0$ and $1/c$ is nonintegral

2 Answers2