Let $X$ be an $L^2$ random variable. Show we can establish the following bound: $$ \mathbb{P}(|X| \geq \gamma \mathbb{E}[|X|]) \geq (1- \gamma)^2\frac{\mathbb E[|X|]^2}{\mathbb{E}[X^2]} $$ for $0< \gamma <1$.
I am thinking this has something to do with Chebyshev's inequality, however I'm not sure. Does anyone have ideas?
Assume, for a contradiction, that for some $\gamma\in (0,1)$, $$ P(|X|>\gamma E(|X|))<(1-\gamma)^2\frac{E[|X|]^2}{E[X^2]} $$ Observe that $$ E[|X|]^2=(E(|X|\mid |X|=0)+E(|X|\mid |X|>0))^2=E(|X|\mid |X|>0)^2 $$ More formally, $$ E|X|=\int_{\Omega}|X|\chi_{|X|>0}dp\leq \|X\|_2\sqrt{P(|X|>0)} $$ by Cauchy-Schwarz-Teichmueller. Hence $$ E[|X|]^2\leq \|X\|_2^2P(|X|>0)=P(|X|>0)E(X^2) $$ Therefore, our assumption yields $$ P(|X|-\gamma E(|X|)>0)<(1-\gamma)^2P(|X|>0) $$ and taking the expectation on both sides, we have $$ E(|X|-\gamma E(|X|))=(1-\gamma)E(|X|)<(1-\gamma)^2E(|X|) $$ which is a contradiction since $\gamma\in (0,1)$.
Hope this helped.