# Integral inequality and the Hölder inequality

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?

• English note: "prove", not "proove". TeX note: only symbols intended to be set in a math font should be set in math mode; for example, "$S$ $\sigma$-algebra" $S$ $\sigma$-algebra (or, even better, "$S$ a $\sigma$-algebra") instead of "‍‍$S\ \sigma-algebra$" $S\ \sigma-algebra$. I have edited accordingly. Commented Mar 12, 2022 at 15:00
• @LSpice. Writing "only symbols intended to be set in a math font should be set in math mode" might be misinterpreted and lead to the other extreme that is often seen where only math symbols are enclosed in $: $\int$f d$\mu$ Commented Mar 12, 2022 at 21:20 • @md2perpe, I meant to include$f$in your example$\int f\,\mathrm d\mu$as a symbol (that happens to be a letter!) that should be set in math mode—and arguably even the$\mathrm d$, as a symbol that should be set in math mode but (I but not everyone believe) using \mathrm—but I agree it could easily be interpreted otherwise. Thank you for the clarification. Commented Mar 12, 2022 at 23:18 • @LSpice. I understood what you meant. I was just a bit worried that your wordings would be misinterpreted. Commented Mar 12, 2022 at 23:27 ## 1 Answer 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. • Is there any way to prove this without using the Jensen's inequality? Commented Mar 12, 2022 at 13:37 • I don't see any other way (which does not mean that there isn't one). At least, I think Cauchy Schwarz is not suitable to prove this inequality since i.g.$\int_X f d\mu < (\int_X f^2 d\mu)^{1/2}$(and the same for$g\$), so you can't go to the inequality you want from there.
– Gerd
Commented Mar 12, 2022 at 14:15
• I was asking because the hint for the solution is to use the Holder inequality, so I tried to use that, but I don't see how it cluld come in handy. Commented Mar 12, 2022 at 14:19
• I edited the answer. Indeed one can use Cauchy Schwarz.
– Gerd
Commented Mar 12, 2022 at 14:26