Assume $X$ is a random variable with probability density function $f(x)=2 x e^{-x^2}$ and $0 \leq L\leq1$ is another random variable with $L\geq1-\frac{1}{4X^2}$. Using the law of total expectation:
\begin{align} E[L]&=E[L|X>\psi] P(X>\psi)+ E[L|X\leq\psi] P(X\leq\psi)\\ &\geq E[L|X>\psi] P(X>\psi)\\ &= E[L|X>\psi] (1- e^{-\psi^2})\\ &=E[1-1/(4X^2)|d>\psi] (1- e^{\psi^2})\\ &\geq (1-1/(4\psi^2)) (1- e^{-\psi^2}) \end{align}
Therefore, with $\psi=\sqrt{2}$, $$E[L]\geq (7/8)(1-e^{-2})>0.$$ I understand from this equation that although $L$ has singularity at $X=0$, since the event $X=0$ occurs with zero probability, the expectation of $L$ remains finite. But, instead of using the law of total expectation, is there a shorter way to prove $E[L]>0$? I mean, by only using the integral for the expectation and considering the fact that the event $X=0$ happens with zero probability. My integral solution is useless because it proves $E[L]>-\infty$:
\begin{align} E[L] &\geq \int\limits_{x=0}^{\infty}(1-1/4x^{2}) 2xe^{-x^2}dx=1-0.5\int\limits_{x=0}^{\infty}\frac{e^{-x^2}}{ x} dx \to -\infty \end{align}
The integral version of my first solution is:
\begin{align} E[L] \geq \int\limits_{x=0}^{\infty}(1-1/4x^{2}) 2xe^{-x^2}dx &\geq \int\limits_{x=\psi>0}^{\infty}(1-1/4x^{2}) 2xe^{-x^2}dx\\ &\geq (1-1/(4\psi^2))\int\limits_{x=\psi>0}^{\infty}2xe^{-x^2}dx\\ &= (1-1/(4\psi^2))(1-e^{-\psi^2})\\ \end{align}
One other way to think about it (according to comments): \begin{align} E[L] &\geq \int\limits_{x=0}^{\infty}\max{(1-1/4x^{2},0)} 2xe^{-x^2}dx\geq 0 \end{align} which is convergent because