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This is for a qualifying exam, so the simpler the solution, the better.

Suppose $\lambda> 0$. I wish to show $$\dfrac{e^{-\lambda}}{(1-e^{-\lambda})^2} \leq \dfrac{1}{\lambda^2}$$

I tried the following: rewrite $e^{-\lambda} = (e^{\lambda/2})^{-2}$ so that we have $$\dfrac{e^{-\lambda}}{(1-e^{-\lambda})^2} = \dfrac{1}{(e^{\lambda/2})^{2}(1-e^{-\lambda})^2} = \dfrac{1}{(e^{\lambda / 2}-e^{-\lambda/2})^2}$$ so this would be equivalent to showing that $$\dfrac{1}{(e^{\lambda / 2}-e^{-\lambda/2})^2} \leq \dfrac{1}{\lambda^2}$$ or $$\lambda^2 \leq (e^{\lambda / 2}-e^{-\lambda/2})^2$$ Nothing too insightful here, unfortunately. If more context is needed, I can provide it.

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    Can you use the inequality $\lambda < e^\lambda$? – 2017-01-14
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    @HenryW. This is a bit of a gray area, as this is for a stats qualifying exam - so we don't approach everything mathematically from first principles. But for the purposes of this question, assume you can. Please note the edit. I made a few typos. – 2017-01-14
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    This is equivalent to showing that $\sinh\lambda\geqslant\lambda$ for every nonnegative $\lambda$, which is direct since the derivative of $\sinh\lambda$ is $\cosh\lambda\geqslant1$. (In addition, one gets that the inequality at the beginning is strict and holds for every real nonzero $\lambda$.) – 2017-01-14

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By multiplying by $e^\lambda$ and clearing denominators, we obtain that the given inequality is equivalent to

$$\lambda^2\leq e^\lambda(e^\lambda-1)^2$$

Now, using the inequality $e^\lambda\geq\lambda+1$, we obtain that

$$e^\lambda(e^\lambda-1)^2\geq(\lambda+1)\lambda^2\geq \lambda^2$$

as desired.

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    No, $\frac{e^{-\lambda}}{(1-e^{-\lambda})^2}$ is not the inverse of $e^\lambda(e^\lambda-1)^2$. – 2017-01-14
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    @upvoter Why the upvote? – 2017-01-14
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Equivalently, for $x>0,$ that $$x^2\leq (e^{-x}-1)^2/e^{-x}= e^{-x}-2+e^x=$$ $$=(1-x/1!+x^2/2!-x^3/3!+...)-2+(1+x+x^2/2!+x^3/3!+...)=$$ $$=2(x^2/2!+x^4/4!+x^6/6!+...)=$$ $$=x^2+2(x^4/4!+x^6/6!+...).$$