# Integral Version of Hölder Inequality

Statement: Let $f_1,f_2,\cdots, f_n$ are $n$ functions who are positive in the region $[a,b]$. And $p_1, p_2, \cdots,p_n$ be positive rational numbers such that $p_1+p_2+\cdots+p_n=1 .$ Then$\prod_{k=1}^n \left[\left(\int_a^b f_k (x)dx \right)^{p_k}\right]\geq \int_a^b\left[ \prod_{k=1}^n\left(f_k (x)\right)^{p_k}\right]dx$

Proof:
$\sum_{k=1}^n \left[\frac{p_kf_k}{\displaystyle{\int_a^b f_k (x)dx}}\right]\geq \prod_{k=1}^n \left[\left(\frac{f_k (x)}{\displaystyle{\int_a^b f_k (x)dx}} \right)^{p_k}\right]$ $\implies \int_a^b \left[\sum_{k=1}^n \left[\frac{p_kf_k(x)}{\displaystyle{\int_a^b f_k (x)dx}}\right] \right]dx \geq \int_a^b \left[ \prod_{k=1}^n \left[\left(\frac{f_k (x)}{\displaystyle{\int_a^b f_k (x)dx}} \right)^{p_k}\right]\right]dx$ $\implies \sum_{k=1}^n \left[\frac{\displaystyle{p_k\int_a^bf_k(x)dx}}{\displaystyle{\int_a^b f_k (x)dx}}\right]\geq \frac{\displaystyle{\int_a^b \left[ \prod\limits_{k=1}^n \left( f_k (x)\right)^{p_k}\right]dx }}{\displaystyle{\prod\limits_{k=1}^n \left[\left(\int_a^b f_k (x)dx \right)^{p_k}\right]}}$ $\implies \prod\limits_{k=1}^n \left[\left(\int_a^b f_k (x)dx \right)^{p_k}\right] \geq \int_a^b\left[ \prod_{k=1}^n\left(f_k (x)\right)^{p_k}\right]dx$ Note by Soham Chatterjee
9 months, 3 weeks ago

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## Comments

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Nice!

- 9 months, 3 weeks ago

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Wonderful I really love the argument

- 9 months, 3 weeks ago

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