Let $0 < \delta < \alpha$, $X$ denote a real random variable and $X_\alpha=X\cdot1_{\{\vert X \vert \leq \alpha\}}$. I want to prove that $$ P(\delta < \vert X \vert \leq \alpha) \leq c \text{Var}[X_\alpha], $$ where $c$ is a constant (which may very possibly depend on $\delta$ and/or $\alpha$).
I have thought about applying Chebyshev's inequality but then you need $X$ to be centered. If $X$ was centered then could do the following computation: $$ P(\delta < \vert X \vert \leq \alpha) \leq P(\delta < \vert X_\alpha \vert \leq \alpha) \leq P( \vert X_\alpha \vert > \delta) \overset{C.I.}{\leq} \delta^{-2}\text{Var}[X_\alpha]. $$ But alas it doesn't appear to be that simple. I have also tried simply writing out $$ \text{Var}[X_\alpha] = \int_\Omega (X_\alpha - \mathbb{E}[X_\alpha])^2 dP = \int_\mathbb{R} (x - \mathbb{E}[X_\alpha])^2 P_{X_\alpha}(dx) = \int_{\{\vert x \vert \leq \alpha\}} (x - \mathbb{E}[X_\alpha])^2 P_{X}(dx) $$ but this doesn't lead to anything useful as far as I can see. Can anyone help me prove this?