The standard proof of Minkowski's inequality in $L^p$ space using Hölder's inequality seems to be pretty unmotivated (see here: http://en.wikipedia.org/wiki/Minkowski%27s_inequality). Breaking up that factor into two parts, and applying Hölder's to each of them separately makes no geometrical intuitive sense to me. Rather, it would appear that the following proof (which I came up with, though is highly probable isn't in fact mine), appears much more motivated:
We need to show that $(\int_{X} |f+g|^p d\mu)^{\frac{1}{p}} \le (\int_{X} |f|^p d\mu)^{\frac{1}{p}} +(\int_{X} |g|^p d\mu)^{\frac{1}{p}}$.
Let $A=(\int_{X} |f|^p d\mu)^{\frac{1}{p}}$, and $B=(\int_{X} |g|^p)^{\frac{1}{p}} d\mu$.
Then setting $f_1=\frac{f}{A}$, and $g_1=\frac{g}{B}$, we get the equivalent inequalities
$\int_{X} (Af_1+Bg_1)^p d\mu \le (A+B)^p$
$\int_{X} (\frac{Af_1+Bg_1}{A+B})^p d\mu \le 1$
$\int_{X} (\frac{Af_1+Bg_1}{A+B})^p d\mu \le \int_{X} \frac{A|f_1|^p+B|g_1|^p}{A+B} d\mu$
Which follows immediately from the convexity of $x^p$.
Now, there are a few reasons that I find this more appealing. First, each of the steps is motivated. The first step is an attempt at homogenization, then after rewriting it in the form with the R.H.S.=1, it becomes very apparent how to finish it off using the convexity of $x^p$.
Second, a similar line of reasoning is used to prove Hölder's inequality, and is used in the following quite elegant proof of Cauchy-Schwarz for finite sequences:
To prove $(\sum a_i^2)(\sum b_i^2) \ge (\sum a_ib_i)^2$, we first homogenize by setting $\sum a_i^2=\sum b_i^2=1$. Then after square rooting, we see it is equivalent to the inequality $\frac{\sum a_i^2 + \sum b_i^2}{2} \ge \sum a_ib_i$, which is a direct result of AM-GM. Note that the use of homogenization is specifically used so that we can replace the L.H.S. with something much stronger while keeping the R.H.S. the same, almost as if by magic (which was essentially what my proof amounted to).
Third, the equality case almost becomes trivial to see, since only one inequality was applied in the entire process, and that was at the very end of the proof. Further, it's very easy to see geometrically the idea of convexity being used here, so the equality case also seems natural.
My first question is, whether there are merits to the standard proof involving Holder's inequality that I seem to be missing which either make it more general, or make it of more interest than the proof I presented.
My second question is to whether the briefly summarized proof involving dual spaces presented on the same Wikipedia page is in actuality more general than my proof. That is, whether there are spaces to which we simply cannot appeal to a convexity argument, and instead have to resort to the supremum proof. Or whether my proof can somehow be modified to prove the result in a more general class of spaces which encompasses all of those which the supremum argument works for.
Cheers,
Rofler