I guess this result is true, so I'm trying to prove it. Could you have a check on my attempt?
Let $X$ be a non-negative bounded random variable. Then $$ \mathbb E[e^{-X}] \le \frac{1}{e\mathbb E X}. $$
Proof: The inequality holds if $\mathbb E X=0$. Now we assume $\mathbb E X>0$. Fix $\varepsilon>0$. Then $$ \mathbb E[e^{-(X+\varepsilon)}] = \mathbb E \left [(X+\varepsilon)e^{-(X+\varepsilon)} \frac{1}{(X+\varepsilon)} \right] \le \frac{1}{e} \mathbb E \frac{1}{(X+\varepsilon)} \le \frac{1}{e} \frac{1}{\mathbb EX + \varepsilon}. $$
The second inequality follows from $\max \{xe^{-x} \mid x \ge 0\} = e^{-1}$. The second one follows from Jensen inequality for the concave function $x \mapsto x^{-1}$. The result then follows by taking the limit $\varepsilon \to 0^+$.
The result does not seem to be true: Take $X$ to be $0$ with probability $0.5$ and $1,000,000$ with probability $0.5$. Then $E[e^{-X}] \geq 0.5$ but $EX \geq 500,000$.