I'm not sure what you mean by $Range(X)$, but assuming you mean the difference between the (essential) sup and inf then the answer is yes. The requirement that $X$ be discrete is also unnecessary.
By translating $X$ we can make the infimum be 0. This doesn't change the variance or range but decreases $\mathbb{E}(X)$ so proving this case is enough. Multiplying then by $\max^{-1}(X)$ we get a r.v. with range of 1 (this multiply both sides of the inequality by a factor of $\max^{-2}(X)$. We now get: $$16var(X)\le 8\mathbb{E}(X)+1 \ .$$ Now, among all r.v. in $[0,1]$ with $\mathbb{E}(X)=e$ the one which maximizes the variance is the one with $\mathbb{P}(X=1)=e$ and $\mathbb{P}(X=0)=1-e$ which has variance of $e-e^2$. So we get the inequality $$16(e-e^2) \le 8e+1$$ for any $0\le e \le 1$ which is equivalent to $$0\le (4e-1)^2 \ .$$