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?
Best Answer
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:
$$4 \left(\sum a_i^4\right)^3 \ge 100^2 \implies \sum a_i^4 \ge \sqrt[3]{2500}$$