Equality harder than inequality

Question

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.

Answer 1

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 $a_1 = a_2 = \cdots = a_n$.

Answer 2

By RMS-AM inequality we know that equality holds if and only iff $a_1=...=a_n$, then we can refer to that proof.

Answer 3

Rewrite the question as $$ \displaystyle \frac1n\sum_{i=1}^{n} a_i \leq \displaystyle \left(\frac1{n} \sum_{i=1}^{n} a_i^{3} \right)^{\frac13} $$ Now this is true, since it is the power mean inequality.

For the power mean inequality (as for all inequalities comparing means), it is well known that equality holds if and only if $a_1=a_2=...=a_n$.

A reference for the last statement.

Answer 4

By Holder $$n^2\sum_{i=1}^na_i^3=\left(\sum_{i=1}^n1\right)^2\sum_{i=1}^na_i^3\geq\left(\sum_{i=1}^n\sqrt[3]{1^2a_i^3}\right)^3=\left(\sum_{i=1}^na_i\right)^3.$$ The equality occurs for $$(1,1,...,1)||(a_1^3,a_2^3,...,a_n^3)$$ or for $$a_1=a_2=...=a_n$$