The typical way I've seen to prove Minkowski's inequality is to take
$$
|f(x) + g(x)|^p \leq |f(x)||f(x)+g(x)|^{p-1} + |g(x)||f(x)+g(x)|^{p-1},
$$
integrate over $x$, and apply Holder's inequality to each term on the right, then divide by $\left( \int|f(x) + g(x)|^p \right)^{(p-1)/p}$.
Suppose we wanted to provide a more direct proof for the finite-dimensional case, and where $p$ is an integer. Let $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ have all non-negative entries, and try to show that
$$
\left(\sum^n (x_i + y_i)^p \right)^{1/p} \leq \left( \sum^n {x_i}^p \right)^{1/p} + \left(\sum^n {y_i}^p \right)^{1/p}. \tag{1}
$$
Raise both sides to the $p$th power and cancel out the ${x_i}^p$ and ${y_i}^p$ terms, and I claim that (1) is true iff
$$
\binom{p}{1}\sum^n {x_i}^{p-1}y_i + \binom{p}{2}\sum^n {x_i}^{p-2}{y_i}^2 + \dotsb + \binom{p}{p-1}\sum^n{x_i}{y_i}^{p-1}\\
\leq \binom{p}{1}\left( \sum^n {x_i}^p \right)^{(p-1)/p}\left( \sum^n {y_i}^p \right)^{1/p} + \binom{p}{2} \left( \sum^n {x_i}^p \right)^{(p-2)/p}\left( \sum^n {y_i}^p \right)^{2/p}\\
+ \dotsb + \binom{p}{p-1}\left( \sum^n {x_i}^p \right)^{1/p}\left( \sum^n {y_i}^p \right)^{(p-1)/p}.
$$
A convenient way to show this would be if the equality held termwise. For the first and last terms, indeed it does, since e.g.
$$
\sum^n{x_i}^{p-1}y_i \leq \left( \sum^n {x_i}^p \right)^{(p-1)/p}\left( \sum^n {y_i}^p \right)^{1/p}
$$
by Holder's inequality. I suspect the same type of inequality holds between the intermediate terms, but I can't see a way to prove it. Some clever application of Holder's inequality, perhaps? For instance, how can we show
$$
\sum^n{x_i}^{p-2}{y_i}^2 \leq \left( \sum^n {x_i}^p \right)^{(p-2)/p}\left( \sum^n {y_i}^p \right)^{2/p}?
$$
Best Answer
A direct proof (not via Hölder's inequality) is not only possible, but is also more natural, as it directly gets to the heart of the matter: convexity. It works the same for integrals and sums.
Let $f=af_0$ and $g=bg_0$ where $f_0,g_0$ are unit vectors. Since the function $u\mapsto |u|^p$ is convex, for $t\in [0,1]$ we have $$|tf_0(x)+(1-t)g_0(x)|^p\le t|f_0(x)|^p+(1-t)|g_0(x)|^p \tag1$$ pointwise. Integrate (or sum) both sides: $$\|tf_0+(1-t)g_0\|_p^p\le t +(1-t) = 1 \tag2$$ With $t=\frac{a}{a+b}$ the above becomes $$\|f+g\|_p^p\le (a+b)^p = (\|f\|_p+\|g\|_p)^p \tag3$$ which is the desired inequality.