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!

  • 1
    $\begingroup$ Sorry, I think I just worked out the solution. It follows from the integral version of Titu's lemma. $\endgroup$ – Usermath Feb 8 at 23:31

Thanks, Usermath for the hint with Titu's lemma, because as I am working on some similar inequalities, I'd like to write it down - just for record.

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 $$ \int\frac{(pf)^2+(qg)^2}{pf+qg} \ge p^2 + q^2$$


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