TITLE: Prove that: $\sqrt[3]{a_1^3+ a_2^3 +\cdots+a_n^3} \le \sqrt{a_1^2 + a_2^2 +\cdots+a_n^2}$
QUESTION [6 upvotes]: Let $a_1, a_2, \ldots, a_n \in \mathbb{R}$. Prove that the following inequality holds:
$$\sqrt[3]{a_1^3+ a_2^3 +\cdots+a_n^3} \le \sqrt{a_1^2 + a_2^2 +\cdots+a_n^2}$$

I first tried to restrict the inequality to $\mathbb{R}^{+}$ and that's fairly easy to do since:
$$\sqrt[3]{a_1^3+ a_2^3 +\cdots+a_n^3} \le \sqrt[3]{|a_1^3| + |a_2^3| +\cdots+ |a_n^3|}$$
and
$$\sqrt{a_1^2 + a_2^2 +\cdots+a_n^2} = \sqrt{|a_1^2| + |a_2^2| +\cdots+ |a_n^2|}$$
I tried using the Jensen's Inequality on $x^{\frac 32}$ and $x^{\frac 23}$, but the inequality goes in the opposite direction. Also I tried to divide with $n^{\frac12}$ and $n^{\frac13}$ and use the power mean inequality, but again the inequality goes in the opposite direction.
Anyway, I managed to solve it by raising the inequality to sixth power and we get:
$$\sum a_1^6 + 3\sum a_1^4a_2^2 + 6\sum a_1^2a_2^2a_3^2 \ge \sum a_1^6 + 2\sum a_1^3a_2^3$$
Which is true since:
$$a_1^4 a_2^2 + a_1^2 a_2^4 \ge 2a_1^3 a_2^3$$
and all terms on the LHS are positive. But this prove looks kind of messy. Is there any better proof?
Also it's seems that we can extend this inequality to:
$$k_1 \ge k_2 \implies \sqrt[k_1]{a_1^{k_1}+ a_2^{k_1} +\cdots+a_n^{k_1}} \ge \sqrt[k_2]{a_1^{k_2} + a_2^{k_2} +\cdots+a_n^{k_2}}$$ But can we actually do that?

REPLY [2 votes]: Since, $\sqrt[3]{\sum\limits_{i=1}^na_i^3}\leq\sqrt[3]{\sum\limits_{i=1}^n|a_i|^3}$ and the right side does not depend on changing sings of our variables,  it's enough to prove our inequality for non-negatives $a_i$. 
Let $f(x)=x^{\frac{2}{3}}$, $a_i^3=x_i$ and $x_1\geq x_2\geq...\geq x_n$.
Thus, we need to prove that
$$\left(\sum_{i=1}^nx_i\right)^{\frac{1}{3}}\leq\left(\sum_{i=1}^nx_i^{\frac{2}{3}}\right)^{\frac{1}{2}}$$ or
$$\sum_{i=1}^nf\left(x_i\right)\geq f\left(\sum_{i=1}^nx_i\right)$$ or
$$\sum_{i=1}^nf\left(x_i\right)\geq f\left(\sum_{i=1}^nx_i\right)+f(0)+...+f(0),$$
which is true by Karamata because $f$ is a concave function and
$$\left(\sum_{i=1}^nx_i,0,...,0\right)\succ(x_1,x_2,...,x_n).$$