[proofplan] We reduce to the case $\|f\|_p = \|g\|_q = 1$ by normalisation, then apply Young's inequality $ab \leq a^p/p + b^q/q$ pointwise and integrate. Young's inequality follows from the concavity of the logarithm. The trivial cases ($\|f\|_p = 0$ or $\|g\|_q = 0$) are handled separately. [/proofplan] [step:Dispose of the trivial cases] If $\|f\|_p = 0$ or $\|g\|_q = 0$, then $f = 0$ or $g = 0$ almost everywhere, so $\int |fg| \, d\mu = 0$ and the inequality holds. If $\|f\|_p = \infty$ or $\|g\|_q = \infty$ and the other norm is positive, the right side is $\infty$ and the inequality is trivial. [/step] [step:Normalise to reduce to unit norms] Assume $0 < \|f\|_p, \|g\|_q < \infty$. Define $\tilde{f} = |f|/\|f\|_p$ and $\tilde{g} = |g|/\|g\|_q$, so $\|\tilde{f}\|_p = \|\tilde{g}\|_q = 1$. It suffices to show $\int \tilde{f}\tilde{g} \, d\mu \leq 1$. [/step] [step:State Young's inequality from the concavity of $\log$] For $a, b \geq 0$ and conjugate exponents $p, q > 1$ (so $1/p + 1/q = 1$), Young's inequality states \begin{align*} ab \leq \frac{a^p}{p} + \frac{b^q}{q}. \end{align*} This follows from the concavity of $\log$: $\log(a^p/p + b^q/q) \geq (1/p)\log(a^p) + (1/q)\log(b^q) = \log(a) + \log(b) = \log(ab)$. [/step] [step:Apply Young's inequality pointwise and integrate to conclude] Applying Young's inequality pointwise with $a = \tilde{f}(x)$ and $b = \tilde{g}(x)$, \begin{align*} \int_E \tilde{f}\tilde{g} \, d\mu \leq \int_E \left(\frac{\tilde{f}^p}{p} + \frac{\tilde{g}^q}{q}\right) d\mu = \frac{1}{p}\|\tilde{f}\|_p^p + \frac{1}{q}\|\tilde{g}\|_q^q = \frac{1}{p} + \frac{1}{q} = 1. \end{align*} Unwinding the normalisation gives $\int |fg| \, d\mu \leq \|f\|_p\|g\|_q$. [/step]