If $Var(X)\leq 1/2$, prove that $P(E(X)-1\leq X\leq 2E(X))\geq 1/2$. 

Let $X$ be a positive valued random variable, 
     such that $Var(X)\leq 1/2$ and $E(X)\in \mathbb{R}$. Prove that
$$P\big(E(X)-1\leq X\leq 2E(X)\big)\geq 1/2.$$


I was thinking of the standard tools, but both Markov, Chebyshev inequalities provide a type $\leq$ inequality.
Thanks in advance for the help. 
 A: Let $\mu=E(X)$ and $\sigma^2=Var(X)$. There are several ways to express Chebyshev's Inequality. For bounded probabilities, that involve an interval of X $\textit{not}$ centered on $\mu$:
$$
P(|X-c|\leq t)\geq 1-\frac{E[(X-c)^2]}{t^2}
$$ 
Now $|X-c|\leq t$ is the same as $-t+c<X<t+c$. If we instead had $a<X<b$, which is the asymmetric, 2-sided case, we can get back to the inequality by setting $a=-t+c$ and $b=t+c$. 
$$
P(a\leq X \leq b)=P\left(\left|X-\frac{a+b}{2}\right|\leq \frac{b-a}{2}\right)\geq 1- \frac{\sigma^2+\left(\mu-\frac{a+b}{2}\right)^2}{\left(\frac{b-a}{2}\right)^2}
$$
For the original problem, set $a=\mu-1>0$ (because X is positive-valued) and $b=2\mu$
$$
P(\mu-1\leq X\leq2\mu)\geq 1- \frac{\sigma^2+\left(\mu-\frac{3\mu-1}{2}\right)^2}{\left(\frac{\mu+1}{2}\right)^2}\geq 1- \frac{1/2+\left(\frac{1-\mu}{2}\right)^2}{\left(\frac{1+\mu}{2}\right)^2}
$$
Now the fraction term in the last inequality is $\leq 1/2$ because we assumed $\mu>1$ without loss of generality (otherwise, if $\mu\leq 1$, then we just truncate the lower bound at zero, and it is even easier to arrive at the final inequality).
Then $P(\mu-1\leq X\leq2\mu)\geq 1/2$.
