If $a_i \in \mathbb R^+$. Prove that $\displaystyle \left( \sum_{i=1}^{n} a_i \right) ^{3} \leq n^2 \displaystyle \left( \sum_{i=1}^{n} a_i^{3} \right)$
Also, show that the equality holds iff
$a_1=a_2=…=a_n$
I have proved the inequality using RMS-AM as well as cauchy-schwarz inequality. But I am stuck in proving the equality condition. Please help.
Best Answer
Let $\bar{a} = \frac1n\sum\limits_{k=1}^n a_k$. Since all $a_k > 0$, so does $\bar{a}$. For each $k$, we have
$$a_k^3 - 3\bar{a}^2a_k + 2\bar{a}^3 = (a_k+2\bar{a})(a_k - \bar{a})^2 \ge 0 $$
Summing over $k$, the sum of LHS becomes
$$\sum_{k=1}^n a_k^3 - 3\bar{a}^2\sum_{k=1}^n a_k + 2n\bar{a}^3 = \sum_{k=1}^n a_k^3 - n\bar{a}^3$$
Multiply the sum on both sides by $n^2$, we obtain
$$n^2\sum_{k=1}^n a_k^3 - \left(\sum_{k=1}^n a_k\right)^3 = n^2 \sum_{k=1}^n (a_k + 2\bar{a})(a_k - \bar{a})^2 \ge 0$$
In order for the inequality to become an equality, the necessary and sufficient condition is $$(a_k + 2\bar{a})(a_k - \bar{a})^2 = 0,\forall k \quad\iff\quad a_k - \bar{a} = 0, \forall k$$ The last condition is equivalent to $a_1 = a_2 = \cdots = a_n$.