let $f$ be a non-negative function on $(-\infty, + \infty)$, satisfying $$ \int_{-\infty}^{+\infty} f(x) \mathrm{d} x=\int_{-\infty}^{+\infty} x^2 f(x) \mathrm{d} x=1, \quad \int_{-\infty}^{+\infty} x f(x) \mathrm{d} x=0, $$ prove that for any $\alpha \leq 0$, $$ \int_{-\infty}^\alpha f(x) \mathrm{d} x \leqslant \frac{1}{1+\alpha^2} . $$
My thought: $\int_R (x-a)^2 f(x) dx = 1+\alpha^2$, which gives $\int_R x^2 f(x+\alpha) dx=1+\alpha^2$. Let $g(x)=f(x+\alpha)$ and then similarily we have $\int_R x^2 g(x) dx = 1+\alpha^2$, $\int_R x g(x) dx = -\alpha$, $\int_R g(x) dx = 1$. Then we only need to prove that $\int_{-\infty}^0 g(x) \leqslant \frac{1}{1+\alpha^2}$.