Integral inequality and the Hölder inequality

Question

Let $\mu:S\rightarrow[0, +\infty]$ be a positive measure on $S$ $\sigma$-algebra on $X$ such that $\mu(X)=1$, and let be $f,g:X\rightarrow \mathbb{R}$ be positive $S$-measurable functions such that: $$f(x)g(x)\geq1$$ $\mu$-almost everywhere in $X$. Prove that: $$\int_X f d\mu\cdot\int_X gd\mu\geq1$$ So, I've proved that $\int_X f(x)g(x)d\mu\geq1$ and then I used the Hölder inequality and obtained: $$\left(\int_X f^2 d\mu\right)^{\frac12}\cdot\left(\int_X g^2d\mu\right)^{\frac12}\geq1.$$ How do I move forward from this point?

Answer 1

As $t \mapsto 1/t$ is convex on $(0,\infty)$ Jensen's inequality gives $$ \frac{1}{\int_X f d\mu} \le \int_X \frac{1}{f} d\mu. $$
From this inequality and $g \ge 1/f$ a.e. we get $$ \int_X f d\mu\int_X g d\mu \ge \int_X f d\mu\int_X \frac{1}{f} d\mu \ge 1 $$

Edit: A way to use Cauchy Schwarz instead of Jensen's inequality: Since $\sqrt{fg} \ge 1$ a.e. we have $$ 1=\int_X 1 d\mu \le \int_X\sqrt{f}\sqrt{g} d\mu \le (\int_X f d\mu)^{1/2} (\int_X g d\mu)^{1/2}. $$ Squaring this inequality leads to the desired result.