Prove that $(a+b)^{\gamma}\ge a^{\gamma} +b^{\gamma}$ whenever $\gamma\ge1$ and $a,b \ge 0$ Prove that $(a+b)^{\gamma}\ge a^{\gamma} +b^{\gamma}$ whenever $\gamma\ge1$ and $a,b \ge 0$. Also, show that  the reverse inequality holds when $0\le\gamma\le1$. (Exercise 1.38 in Stein's Real Analysis)
[Hint: Integrate the inequality between $(a + t)^{\gamma-1}$ and   $ t^{\gamma-1}$ from 0 to b.]
I can only prove it for integer $\gamma$ and I don't know how to use the hint. Besides, what is the relation between this exercise and the content of this chapter (I mean, measure)?
 A: If $x\ge 0, r\ge 1$, then
$$f(x)=(1+x)^r-x^r \implies f'(x)= r[(1+x)^{r-1}-x^{r-1}]$$ $$ \implies f'(x)=r[(1+1/x)^{r-1}-1]>0.$$ Hence $f(x)$ ia an strictly increasing function for $x \ge 0$ So $$f (x)=(1+x)^r-x^r \ge f(0)=1.$$ Let us take $x=a/b$, then
$$(1+a/b)^n-(a/b)^n>1 \implies (a+b)^r \ge a^r+b^r.$$ 
A: If $0<\gamma\le 1$ then $(a+t)^{\gamma-1}\le t^{\gamma-1}$ for $a$, $t\ge0$, so
that
$$\int_0^b(a+t)^{\gamma-1}\,dt\le\int_0^bt^{\gamma-1}\,dt,$$
that is
$$\frac{(a+b)^\gamma-a^\gamma}{\gamma}\le\frac{b^\gamma}\gamma.$$
When $\gamma\ge1$ you can flip the inequalities.
A: Let $\gamma \geq 1$. $\int_0^{b} (a+t)^{\gamma-1}dt \geq \int_0^{b} t^{\gamma-1}dt$ because $a+t \geq t$ for all $t$. Evaluating the integrals we get $\frac {(a+b)^{\gamma} -a^{\gamma}} {\gamma} \geq \frac {b^{\gamma}} {\gamma}$. Now multiply by $\gamma$ and transfer $a^{\gamma}$ to the right side. The argument is similar when $\gamma <1$.
A: There is another proof to the inequality.
It is suffices to prove 
$$\left(1+\frac{b}{a}\right)^\gamma \geq 1+\left(\frac{b}{a}\right)^\gamma.$$
In particular, for any $\,\gamma>1$, let
$$f(t):=(1+t)^\gamma-(1+t^\gamma).$$
Differentiate directly yields
$$f’(t)=\gamma\left((1+t)^{\gamma-1}-t^{\gamma-1}\right).$$
Since $\gamma>1$, we know$\,f’(t)>0.$ That is $f$ is increasing on t>0.
Note that $f(0)=0$, we get $f(t)\geq 0,\, $for all $t\geq 0$.
Then set $t=\frac{b}{a}$, we are done.
