Inequality of a random variable

Question

I am reading a paper, they let a random variable $a$ and its average $\bar{a}$ satisfy $0 \leq a, \bar{a} \leq 1$, after that, they conclude

$\mathbb{E} \{ (a - \bar{a})(a + \bar{a}) \} \leq 2 \mathbb{E} \{ |a - \bar{a} |\}$

The above inequality may be trivial, but it is not for me. Could you please explain why? Thank you very much!

Answer 1

$$E[(a-\bar{a})(a+\bar{a})]\leq E[|a-\bar{a}|(a+\bar{a})]\leq 2E[|a-\bar{a}|],$$

where the second inequality follows from the fact that $(a+\bar{a})\leq 2$.

Answer 2

As $0 \leq a, \bar{a} \leq 1$, we have

$$(a-\bar{a})(a + \bar{a}) \leq |(a-\bar{a})(a+\bar{a})| = |a-\bar{a}| \cdot \underbrace{|a+\bar{a}|}_{= a+\bar{a} \leq 2}.$$

Taking expectation on both sides, proves the inequality.