Proving an inequality using the Cauchy-Schwarz inequality Consider four real numbers $a_1, a_2, a_3, a_4$ such that $\sum a_i^3 = 10$. Prove that 
$$\sum a_i^4 \geq \sqrt[3]{2500}$$
Applying the Cauchy Schwarz inequality with $a_i^2$ and $a_i$, we get
$$\left(\sum a_i^3\right)^2 \leq \left(\sum a_i^4\right)\left(\sum a_i^2\right)$$
Again, applying the Cauchy Schwarz inequality with $a_i^2$ and $1$, we get:
$$\left(\sum a_i^2\right)^2 \leq 4\left(\sum a_i^4\right)$$
Substituting this into the first inequality, we get:
$$\left(\sum a_i^3\right)^2 \leq 4\left(\sum a_i^4\right)^2$$
Taking the square root of both sides, 
$$\left(\sum a_i^3\right) \leq 2\left(\sum a_i^4\right)$$
$$\implies 5 \leq \sum a_i^4$$
But, $\sqrt[3]{2500} = 13.5720881$, which is more than $2$ times $5$. How do I prove the required statement?
 A: Using the following relations from Cauchy-Schwarz (which has the advantage of being applicable for all reals):
$$\left(\sum a_i^4\right)\left(\sum 1\right)\ge \left(\sum a_i^2\right)^2 \tag{1}$$
$$\left(\sum a_i^4\right)\left(\sum a_i^2\right)\ge \left(\sum a_i^3\right)^2 = 100\tag{2}$$
Noting both LHS and RHS are positive, square both sides of (2) and multiply with (1), then cancel off the positive $(\sum a_i^2)^2 $ from both sides to get:
$$4 \left(\sum a_i^4\right)^3 \ge 100^2 \implies \sum a_i^4 \ge \sqrt[3]{2500}$$
A: I don't think the Cauchy-Schwarz inequality gives a quick proof. If you don't insist on that, the quick proof follows directly from the following inequality:
$$\sqrt[3]\frac{\sum a_i^3}4\leq\sqrt[4]\frac{\sum a_i^4}4$$
That is,
$$\sum a_i^4\geq4\left(\frac{\sum a_i^3}{4}\right)^{\frac43}=\sqrt[3]{2500}$$


Please let me explain something because some mentioned that power mean inequality holds only for non-negative $a_i$. But I insist that if $M_3(a_i)>0, M_4(a_i)>0$, then we can still apply the inequality.

Let us separate $a_i$ according to whether it's non-negative or negative. We can write
$$\sum a_i=\sum a_++\sum a_-\tag{*}$$
then still consider the same inequality
$$\begin{align}\sqrt[3]\frac{\sum a_i^3}4&=\sqrt[3]\frac{\sum a_+^3+\sum a_-^3}4\\
&<\sqrt[3]\frac{\sum a_+^3+\sum(-a_-)^3}4\\
&\leq\sqrt[3]\frac{\sum a_+^4+\sum(-a_-)^4}4\\
&=\sqrt[4]\frac{\sum a_+^4+\sum a_-^4}4=\sqrt[4]\frac{\sum a_i^4}4\end{align}$$
The first and the last line is a simple expansion and collection using (*). The second line follows that negative is always less than positive. And the third line is the direct result of power mean inequality.
Note the second line is strictly less than, which implies if "=" attains, all $a_i$ must be positive.
