Prove that: $\sqrt[3]{a_1^3+ a_2^3 +\cdots+a_n^3} \le \sqrt{a_1^2 + a_2^2 +\cdots+a_n^2}$

This inequality follows easily by induction from the following fact (with $p=2/3$):

$(x+y)^p \leq x^p + y^p$ for all $x,y \geq 0$ and $0 < p < 1$.

To prove this, fix $y$, and remark that the function $x \mapsto (x+y)^p-x^p$ is non-increasing, hence is bounded from above by its value at $x=0$.

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 $$\left(\sum_{i=1}^nx_i,0,...,0\right)\succ(x_1,x_2,...,x_n).$$