Is there a result like $(a_1+...+a_n)^\alpha \leq K(a_1^\alpha +...+a_n^\alpha )$? Is there a result like $$(a_1+...+a_n)^\alpha \leq K(a_1^\alpha +...+a_n^\alpha )$$ 
where $\alpha \geq 1$ real ? I know that for $\alpha =2$ it's true, but how is it in general ?
 A: Hint
Show that if $f$ is convex, then $$f\left(\frac{x_1+...+x_n}{n}\right)\leq \frac{1}{n}\left(f(x_1)+...+f(x_n)\right),$$
whenever $n\geq 2$.
A: By Power Mean inequality for all $\alpha\geq1$ and $a_i>0$ we have:
$$\frac{a_1^{\alpha}+a_2^{\alpha}+...+a_n^{\alpha}}{n}\geq\left(\frac{a_1+a_2+...+a_n}{n}\right)^{\alpha}$$
A: Suppose $f,g:\mathbb R^n\setminus \{0\} \to (0,\infty)$ are continuous and homogeneous of degree $d\in \mathbb R.$ Then there exists a constant $C$ such that $f(x) \le Cg(x)$ for all $x\in R^n\setminus \{0\}.$ (In fact both functions lie between constant multiples of $|x|^d.$) Apply this in the problem at hand with $f(x) = (|x_1| + \cdots + |x_n|)^\alpha,$ $g(x) = |x_1|^\alpha + \cdots+ |x_n|^\alpha,$ and $d=\alpha.$
A: Suppose $a_i>0$. Let $x_i=\frac{a_i}{\sum_{k=1}^na_k}$. Then
$$ \frac{a_1^\alpha+a_2^\alpha+\cdots+a_n^{\alpha}}{(a_1+a_2+\cdots+a_n)^\alpha} =\sum_{i=1}^nx_i^\alpha$$
where 
$$ \sum_{i=1}^nx_i=1,x_i>0.$$
The function $f(x)=\sum_{i=1}^nx_i^\alpha$ reaches the minimum $K=\frac{1}{n^{\alpha-1}}$ under the restriction $\sum_{i=1}^nx_i=1,x_i>0$. So 
$\sum_{i=1}^nx_i^\alpha\ge K$ or
$$ (a_1+a_2+\cdots+a_n)^\alpha \le K(a_1^\alpha+a_2^\alpha+\cdots+a_n^{\alpha}).$$ 
