How can we prove $(a_1+a_2+\cdots+a_n)^r\le a_1^r+ a_2^r+\cdots+a_n^r$ where each $a_i>0$ and $0I am reading  the book: Inequalities by Hardy, Littlewood and Polya.
There are two inequalities at the page number 32:
(1) $(a_1+a_2+\cdots+a_n)^r\le a_1^r+ a_2^r+\cdots+a_n^r$  where each $a_i>0$ and $0<r<1$.
(2) $(a_1+a_2+\cdots+a_n)^r\ge a_1^r+ a_2^r+\cdots+a_n^r$  where each $a_i>0$ and $r>1$.
How can we prove these inequalities. ? Many thanks for your help.
 A: For (1): set $a_i = A_i^{1/r}$, $p=1/r>1$. Then the inequality states
$$ \|A\|_{p} \le \|A\|_1,$$
where $$\|A\|_p := \sqrt[p]{\sum_{i=1}^n A_i^p}.$$
Now have a look here - $\ell^p\subseteq\ell^q$ for $0<p<q<\infty$ and $\|\cdot\|_q<\|\cdot\|_p$
For (2): take $r$th roots of both sides and use (1).
A: Hint: You can use induction. For the initial step of induction, you need to show the (almost trivial) fact that $f(x) = (x+1)^r-x^r$ is an increasing function on $[1,\infty)$ if and only if $r>1$. 
A: By induction: discard any $a_i$ that are zero, since they clearly don't do anything. The basis case is clear. Then
$$ (a_1 + \dotsb + a_n + a_{n+1} )^r = (a_1 + \dotsb + a_n )^r \left( 1 + \frac{a_{n+1}}{a_1 + \dotsb + a_n} \right)^r $$
Now apply Bernoulli's inequality,
$$ (1+x)^r \gtreqless 1+rx $$
if $0 < r \lesseqgtr 1$, the upper/lower inequality symbols being taken together. Then
$$ (a_1 + \dotsb + a_n + a_{n+1} )^r \gtreqless (a_1 + \dotsb + a_n )^r\left( 1+r \frac{a_{n+1}^r}{(a_1 + \dotsb + a_n )^r} \right) = (a_1 + \dotsb + a_n )^r + a_{n+1}^r . $$
The result now follows by applying the induction hypothesis.
A: Let $a_1\geq a_2\geq ...\geq a_n$ and $f(x)=x^{r}$.
Thus, $$(a_1+a_2+...+a_n,0,...,0)\succ(a_1,a_2,...,a_n)$$ and since $f$ is a convex function for $r>1$, by Karamata we obtain:
$$f(a_1+a_2+...+a_n)+f(0)+...+f(0)\geq f(a_1)+f(a_2)+...+f(a_n),$$ which gives a second inequality.
The first inequality we can prove by the same way.
