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?

Answer 1

$(1+t)^{c}-1-t^{c}$ 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 $a^{c}$.

Answer 2

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