\section{Product Inequalities}
\label{section:product-inequalities}

\begin{lemma}[Young's Inequality]
\label{lemma:young-inequality}
    Let $p, q \in (1, \infty)$ such that $1/p + 1/q = 1$, then for any $a, b > 0$,
    \[
    ab \le \frac{a^p}{p} + \frac{b^q}{q}
    \]
    and for any $\eps > 0$,
    \[
    ab \le \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
    \]
\end{lemma}
\begin{proof}
    Since $x \mapsto \exp(x)$ is convex,
    \begin{align*}
        ab &= \exp(\ln(a) + \ln(b)) = \exp\paren{\frac{1}{p}\ln(a^p) + \frac{1}{q}\ln(b^q)} \\
        &\le \frac{1}{p}\exp(\ln(a^p)) + \frac{1}{q}\exp(\ln(a^q)) = \frac{a^p}{p} + \frac{b^q}{q}
    \end{align*}
    For any $\eps > 0$, applying the above inequality to $(\eps p)^{1/p}a$ and $b/(\eps p)^{1/p}$ yields
    \[
    ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}} \le \eps a^p + \frac{b^q}{q}(\eps p)^{-(1/p)q} = \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q
    \]
\end{proof}
