Minkowski inequality $-$ Why should it follow from Holder's?

Solution 1:

The significance of $\frac{1}{p}+\frac{1}{q}=1$ is that this is related to an interpolation. For the exponential, $$ \exp\left( \frac{1}{p}u + \frac{1}{q}v\right) \le \frac{1}{p}e^{u}+\frac{1}{q}e^{v}. $$ Letting $u=\ln(x^p)$, $v=\ln(y^q)$ for $x > 0$ and $y > 0$ gives $$ xy \le \frac{1}{p}x^{p}+\frac{1}{q}y^q. $$ Therefore, if $f\in L^p$ and $g\in L^q$, then $fg\in L^1$ with $$ \int |fg|d\mu \le \frac{1}{p}\|f\|_p^p+\frac{1}{q}\|g\|_q^q \\ \int \frac{|fg|}{\|f\|_p\|g\|_q}d\mu \le \frac{1}{p}+\frac{1}{q}=1 \\ \int |fg|d\mu \le \|f\|_p\|g\|_q. $$ Because of this interpolation, $$ |f+g| = |f+g|^\frac{1}{p}|f+g|^\frac{1}{q}=|f+g|^{\frac{1}{p}}|f+g|^{1-\frac{1}{p}} \\ |f+g|^{p} = |f+g||f+g|^{p-1} \le |f+g|^{p-1}|f|+|f+g|^{p-1}|g| % \\ % (p-1)q = (p-1)\frac{1}{1-1/p}=p $$ That's the reason for the split $|f+g|^{p}=|f+g|^{p-1}|f+g|$, and why you might expect it to yield something useful.