Trouble with Holder Inequality Gymnastics: $\left(\int fg^{q}\right)^{r-p}\leq\left(\int fg^p\right)^{r-q}\left(\int fg^r\right)^{q-p}.$

Question

$\textbf{The Problem:}$ Let $f$ and $g$ be nonnegative and measurable and $0<p<q<r<\infty$. Prove that $$\left(\int fg^{q}\right)^{r-p}\leq\left(\int fg^p\right)^{r-q}\left(\int fg^r\right)^{q-p}.$$

$\textbf{My Thoughts:}$ I noted that $$\frac{r-q}{r-p}+\frac{q-p}{r-p}=1,$$ and since $$\frac{r-p}{r-q},\frac{r-p}{q-p}>1,$$ these are conjugate exponents, and we may use Holder's Inequality. Next, I realized after some trial and error that $$\frac{p(r-q)}{r-p}+\frac{r(q-p)}{r-p}=q.$$ With this in mind, Holder's Inequality implies that \begin{align*} \Large\int fg^q &= \Large\int f^{\frac{r-q}{r-p}}g^{p\frac{r-q}{r-p}}f^{\frac{q-p}{r-p}}g^{r\frac{q-p}{r-p}}\\ &\le \Large\left(\int fg^q\right)^{\frac{r-q}{r-p}}\left(\int fg^r\right)^{\frac{q-p}{r-p}}, \end{align*} and taking $(r-p)$th root yields the inequality.

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Do you agree with my Holder Inequality gymnastics?

Thank you for any feedback, and most important, thank you for your time, I really appreciate it.

Answer 1

Yes, I agree with your Holder Inequality gymnastics.

You are most welcome.