The excellent book The Cauchy-Schwarz Master Class has already been mentioned in the comments by Theo.
Since I cannot stand open questions where a lot of people know the answer to I'll just summarize what is in chapter 9 of the referred book.
You're right that you can prove it the way you do but usually when people take a course in measure and integration theory first Hölder's inequality is proven and then Minkowski's inequality. In that way it can be instructive to use Hölder's inequality to prove Minkowksi's inequality.
There is another advantage to this approach. We can quite easily deduce from the proof when equality arises. To see this let me quickly recall how the proof goes (this can be found in the book by Steele).
First write by using the triangle inequality
$$\sum_{k = 1}^n |x_k + y_k|^p \leq \sum_{k = 1}^n |x_k||x_k + y_k|^{p - 1} + \sum_{k = 1}^n |x_k||x_k + y_k|^{p - 1}.$$
So now we can assume $p > 1$ otherwise we are done. We can now apply Hölder to both of the terms on the right hand side so we find
$$\sum_{k = 1}^n |x_k||x_k + y_k|^{p - 1} \leq \left (\sum_{k = 1} |x_k|^p \right )^{1/p} \left (\sum_{k = 1}^n |x_k + y_k|^{p} \right )^{(p - 1)/p}$$
and
$$\sum_{k = 1}^n |y_k||x_k + y_k|^{p - 1} \leq \left (\sum_{k = 1} |y_k|^p \right )^{1/p} \left (\sum_{k = 1}^n |x_k + y_k|^{p} \right )^{(p - 1)/p}$$
Now we can assume that the left hand side of the first inequality is non-zero so we can divide by $\displaystyle \left (\sum_{k = 1}^n |x_k + y_k|^{p} \right )^{(p - 1)/p}$ to obtain the proof.
Fine. So now if we would have equality in Minkowski's inequality the first inequality written here would also be an equality. This implies $|x_k + y_k| = |x_k| + |y_k|$ for all $1 \leq k \leq n$. Thinking for a bit we can conclude that $x_k$ and $y_k$ must be of the same sign for all $k$. Actually, there is no problem to assume $x_k, y_k \geq 0$ because we can factor the - out and it gets lost in the absolute value.
But equality in Minkowksi's inequality also means that we have equality in the two lines where Hölder's inequality is used. Now you can recall what it means to have equality in Hölder's inequality. We have that there exists $\lambda, \lambda' \geq 1$ such that
$$\lambda |x_k|^p = (|x_k + y_k|^{p - 1})^q = |x_k + y_k|^p \text{ and } \lambda' |y_k|^p = (|x_k + y_k|^{p - 1})^q = |x_k + y_k|^p.$$
Dividing both equalities we get that $\frac{\lambda}{\lambda'} |x_k|^p = |y_k|^p$. So this proof can be easily backtraced.
But again, credit must be given where credit is due: This is just what is written in Steele in my own words. I don't think I'm plagiarizing because this method can be considered to be common knowledge.
So check out the book in the library or buy it, it is quite cheap for a math book and it contains fun exercises.
On leo's request I'm posting my comment as an answer.
Your treatment of the equality cases of Hölder's and Minkowski's inequalities are perfectly fine and clean. There's a small typo when you write that $\int|fg| = \|f\|_p\|g\|_q$ if and only if $|f|^p$ is a constant times of $|g|^q$ almost everywhere (you write the $p$-norm of $f$ and the $q$-norm of $g$ instead).
The case where either one $\|f\|_p$ or $\|g\|_q$ (or both) are infinite isn't part of this exercise and simply wrong. You can trisect $E = F \cup G \cup H$ into disjoint measurable sets of positive measure, take $f$ not $p$-integrable on $F$ and zero on $G$, take $g$ not $q$-integrable on $G$ and zero on $F$ and choose $fg$ non-integrable on $H$. Then certainly no power of $|f|$ is a constant multiple of a power of $|g|$ and vice versa, even though equality holds in the Hölder inequality.
A very nice “blackboard summary” of the equality case (for finite sequences) is given in Steele's excellent book The Cauchy–Schwarz Master Class. Let $a = (a_1,\ldots,a_n) \geq 0$ and $b = (b_1, \ldots, b_n) \geq 0$ and let $\hat{a}_i = \dfrac{a_i}{\|a\|_p}$ and $\hat{b}_i = \dfrac{b_i}{\|b\|_q}$. Then your argument is subsumed by the diagram (with an unfortunate typo in the upper right corner—no $p$th and $q$th roots there):
Mimicking this for functions, let us write $\hat{f} = \dfrac{|f|}{\|f\|_p}$ and $\hat{g} = \dfrac{|g|}{\|g\|_q}$ (assuming of course $\|f\|_p \neq 0 \neq \|g\|_q$), so $\int \hat{f}\vphantom{f}^p = 1$ and $\int \hat{g}^q =1$ and thus your argument becomes
$$
\begin{array}{ccc}
\int |fg| = \left(\int|f|^p\right)^{1/p} \left(\int|g|^q\right)^{1/q} & & |f|^p = |g|^q \frac{\|f\|_{p}^p}{\|g\|_{q}^q} \text{ a.e.}\\
\Updownarrow\vphantom{\int_{a}^b} & & \Updownarrow \\
\int \hat{f}\,\hat{g} = 1 & & \hat{f}\vphantom{f}^p = \hat{g}^q \text{ a.e.} \\
\Updownarrow\vphantom{\int_{a}^b} & & \Updownarrow \\
\int \hat{f}\,\hat{g} = \frac{1}{p} \int \hat{f}\vphantom{f}^p + \frac{1}{q} \int \hat{g}^q & \qquad \iff \qquad &
\hat{f}\,\hat{g} = \frac{1}{p} \hat{f}\vphantom{f}^p + \frac{1}{q} \hat{g}^q \text{ a.e.}
\end{array}
$$
I suggest that you draw a similar diagram for the equality case of Minkowski's inequality.
Best Answer
Let $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$ be two elements of $\mathbb{R}^n_{+}$. Using the concavity of the function $f$ defined by $f(x)=x^p$, $0<p<1$, we have for any $t\in (0,1)$: $$(x_k+y_k)^p=\bigg(t\dfrac{x_k}{t}+(1-t)\dfrac{y_k}{1-t}\bigg)^p\geq t\dfrac{x_k^p}{t^p}+(1-t)\dfrac{y_k^p}{(1-t)^p}.$$ Taking the sum, we obtain $$||x+y||_p^p\geq t\dfrac{||x||_p^p}{t^p}+(1-t)\dfrac{||y||_p^p}{(1-t)^p}.$$ If $t=\dfrac{||x||_p}{||x||_p+||y||_p}$ then $1-t=\dfrac{||y||_p}{||x||_p+||y||_p}$ and we have
\begin{eqnarray*} ||x+y||_p^p &\geq& t\dfrac{||x||_p^p}{\dfrac{||x||_p^p}{\bigg(||x||_p+||y||_p\bigg)^p}}+(1-t)\dfrac{||y||_p^p}{\dfrac{||y||_p^p}{\bigg(||x||_p+||y||_p\bigg)^p}}\\ \Rightarrow ||x+y||_p^p &\geq& t\bigg(||x||_p+||y||_p\bigg)^p+(1-t)\bigg(||x||_p+||y||_p\bigg)^p=\bigg(||x||_p+||y||_p\bigg)^p\\ \Rightarrow||x+y||_p &\geq& ||x||_p+||y||_p. \end{eqnarray*}