Showing $E[X] \geq P(X\geq 2)$ for a random variable under certain conditions 
Let $X$ be a random variable with finite expectation, and suppose that $E[e^{-X}] \leq 1$. Prove that $E[X] \geq P(X\geq 2)$.

My first thought was to use Markov's inequality, but I can only use that if $X$ is non-negative, and in any case, the resulting inequality doesn't seem very helpful. Even if I assume that $X\geq -c$ and apply it to $Y = X+c$, I'm left to show $P(X\geq 2) \geq \frac{c}{c+1}$, which doesn't seem like it's always true.
I'm primarily unsure of how to use the $E[e^{-X}]$ condition; Jensen's inequality tells me that the expectation  is non-negative but there's presumably more to it.
Additionally, I tried rewriting the inequalities in terms of integrals and playing around with parts, but that didn't seem to lead anywhere.
 A: I'm not entirely sure how the hypothesis about the expectation comes into play, but here's my solution. I'll simply use the Markov inequality for a non-negative r.v. and $a > 0$ in the form
$$ E[X] \ge aP(X \ge a) $$
with $a = 2$. Remembering that probabilities are non-negative and using Markov we have:
$$ E[X] \ge P(X \ge 2) + P(X \ge 2) \ge P(X \ge 2) $$
--- Edit after the comments about non-negativity:
I think I saw the light :D I use the extended form of Markov inequality for monotonically increasing function:
Theorem
Let $\varphi$ be a monotonically increasing function for $x > 0$ and let X be a r.v., $a>0, \varphi(a) > 0$. Then
$$ P(X \ge a) \le \frac{\mathbb{E}[\varphi(X)]}{\varphi(a)}$$
In our case I'll consider the (somehwat artificial) function $\varphi(x):= x + e^{-x}-1$, which we can prove satisfy all the hypothesis with $a = 2$. Hence, since $1/\varphi(a) \le 1$ and using the bound for $e^{-X}$, we have
$$ \begin{align*} P(X \ge a) &\le \frac{\mathbb{E}[X+e^{-X}-1]}{\varphi(a)}  \\ &\le \mathbb{E}[X] + \mathbb{E}[e^{-X}] -1 \\ &\le \mathbb{E}[X] \end{align*}$$
