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====== Minkovski's inequality ======

Let $f, g: \Xset \to \rset$ be two measurable functions on a measurable space $(\Xset, \Xsigma)$, and let $\mu$ be a non-negative measure on $(\Xset, \Xsigma)$. Then, for any $p \geq 1$, $$\lr{\int |f+g|^p \rmd \mu}^{1/p} \leq \lr{\int |f|^p \rmd \mu}^{1/p} + \lr{\int |g|^p \rmd \mu}^{1/p}.$$

===== Proof =====
Without loss of generality, we assume that $f, g \in \lp{p}(\mu)$ and $f, g \geq 0$ (the general case can be handled using the inequality $|f+g| \leq |f| + |g|$). For $s > 0$, define $$\varphi(s) = \lr{\int (f + s g)^p \rmd \mu}^{1/p}.$$ Then, the derivative of $\varphi$ with respect to $s$ is $$\varphi'(s) = \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \int (f + s g)^{p-1} g \, \rmd \mu.$$ Using Hölder’s inequality, we can bound $\varphi'(s)$ as follows: $$\varphi'(s) \leq \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \lr{\int (f + s g)^p \rmd \mu}^{\frac{p-1}{p}} \lr{\int g^p \rmd \mu}^{1/p} = \lr{\int g^p \rmd \mu}^{1/p}.$$ Integrating $\varphi'(s)$ from $0$ to $1$, we obtain $$\lr{\int (f + g)^p \rmd \mu}^{1/p} - \lr{\int f^p \rmd \mu}^{1/p} = \varphi(1) - \varphi(0) = \int_0^1 \varphi'(s) \, \rmd s \leq \lr{\int g^p \rmd \mu}^{1/p}.$$ This completes the proof. ◻

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