# An inequality involving two probability densities

I cannot prove the following inequality, which I state below:

Let $$p, q$$ be two positive real numbers such that $$p+q=1$$. Let $$f$$ and $$g$$ be two probability density functions. Then, show that:

$$\int_{\mathbb{R}} \frac{p^2 f^2 + q^2 g^2}{pf + qg} \geq p^2+q^2~.$$

I tried to use Cauchy-Schwarz and even Titu's lemma, but got nowhere. Any help will be greatly appreciated. Thanks!

• Sorry, I think I just worked out the solution. It follows from the integral version of Titu's lemma. – Usermath Feb 8 at 23:31

Titu's Lemma states that for positive reals $$u_i,v_i \in \mathbb{R^+} (i=1,..,n)$$ the following inequality holds: $$\frac{(\sum_{i=1}^n u_i)^2}{\sum_{i=1}^nv_i} \le \sum_{i=1}^n\frac{u_i^2}{v_i}$$ which leads to the integral inequality for suitable non-negative functions $$u,v$$ $$\frac{(\int u)^2}{\int v} \le \int\frac{u^2}{v}$$ Let us set $$u:=pf$$ and $$v=pf+qg$$ then we get together with $$f,g$$ being probability densities $$\int\frac{(pf)^2}{pf+qg} \ge \frac{(\int pf)^2}{\int pf+qg} \\ =\frac{p^2 (\int f)^2}{p\int f+q\int g} \\ =\frac{p^2 }{1} \\ = p^2$$ Swapping the roles of $$pf$$ with $$qg$$ in the inequality above renders $$\int\frac{(qg)^2}{pf+qg} \ge q^2$$ Combining both inequalities we have $$\int\frac{(pf)^2+(qg)^2}{pf+qg} \ge p^2 + q^2$$