Here we shall discuss the proofs of some inequalities. Actually,
we are interested in just one inequality, the last one. We shall
need the first two to prove the last one.
Let $p\in[1, \infty)$ be any real number.
Consider this function from ${\mathbb R}^n\rightarrow{\mathbb R}$
$$
\|\vec a \|_p= \left(\sum_{i=1}^n |a_i|^p \right)^{1/p}.
$$
As you may guess, We are just imitating the definition of
Euclidean norm of $\vec a.$ Our hope is that this is also a
norm. The first conditions of being a norm are easy to check. The
third condition (triangle inequality) is what we shall attack
here. In this context triangle inequality is called Minkowski
inequality. The proof, I am sorry to admit, is not very
intuitive or elegant (in my humble opinion). We start with
another inequality first.
We already have a $p\in[1, \infty).$ Define $q =
p/(p-1).$ Clearly, $q\in[1, \infty)$
and $\frac1p+\frac1q=1.$ Young's inequality says:
If $a,b>0,$ then
$$\frac{a^p}{p}+\frac{b^q}{q} \geq ab.$$
Looks pretty wierd! But the proof is simple.
The LHS is a sum and the RHS is a product. So taking
logarithm naturally comes to mind. The log of the LHS is
$$
\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right).
$$
Notice that, since $\frac1p+\frac1q=1,$ so the LHS was just
a convex combination of $a^p$ and $b^q.$ Since $\log
x$ is a concave function, we must have
$$
\log\left(\frac{a^p}{p}+\frac{b^q}{q}\right) \geq \frac{\log a^p}{p} +
\frac{\log b^q}{q}=\log a + \log b = \log (ab).
$$
Hence the inequality is proved (since $\log x$ is strictly
increasing).
This is the second step in proving the triangle inequality (or
Minkowski inequality). We are still working with the
same $p,q$ as above.
Holder's inequality says: If $\vec a,\vec b\in{\mathbb R}^n$ then
$$
\|\vec a\|_p\|\vec b\|_q\geq \sum|a_ib_i|.
$$
Rather complicated looking! Let us simplify it a bit before trying
the prove it. Divide both sides by the LHS (if the LHS is zero,
then the result is trivial, anyway) to get
$$
1 \geq \sum A_i B_i,
$$
where $A_i = |a_i|/\|\vec a\|_p$ and $B_i = |b_i|/\|\vec b\|_q.$
Just apply Young's inequality to get
$$
A_iB_i \leq \frac{A_i^p}{p}+\frac{B_i^q}{q}.
$$
Summing both sides over $i,$ the RHS
becomes $\frac1p+\frac1q=1,$ completing the proof.
Now at last we in a position to proof Minkowski inequality, which
is just the triangle inequality for $\|\cdots\|_p:$
$$
\|\vec a+\vec b\|_p \leq \|\vec a\|_p+\|\vec b\|_p.
$$
The proof is somewhat indirect. We start with the LHS raised to
the power $p$:
$$
\|\vec a+\vec b\|_p^p = \sum |a_i+b_i|^p= \sum |a_i+b_i|\cdot|a_i+b_i|^{p-1}.
$$
Now apply triangle inequality to only the $|a_i+b_i|$ part
to get
$$
\sum |a_i+b_i|\cdot|a_i+b_i|^{p-1} \leq \sum
(|a_i|+|b_i|)\cdot|a_i+b_i|^{p-1} = \sum|a_i|\cdot|a_i+b_i|^{p-1}
+\sum |b_i|\cdot|a_i+b_i|^{p-1}.
$$
Next, we shall apply Holder's inequality to each of these
terms. For this we shall introduce $q$ as earlier. Also, it
might help to write $c_i = (a_i+b_i)^{p-1}.$
For
example, application of Holder's inequality on the first term
gives
$$
\sum|a_ic_i| \leq \|\vec a\|_p\|\vec c\|_q.
$$
Similarly, the second term gives
$$
\sum|b_ic_i| \leq \|\vec b\|_p\|\vec c\|_q.
$$
So we get
$$
\sum|a_i|\cdot|a_i+b_i|^{p-1}
+\sum |b_i|\cdot|a_i+b_i|^{p-1} \leq (\|\vec a\|_p+\|\vec
b\|_p)\cdot\|\vec c\|_q.
$$
Thus we have got
$$
\|\vec a+\vec b\|_p^p \leq (\|\vec a\|_p+\|\vec
b\|_p)\cdot\|\vec c\|_q.
$$
The result now follows on slight rearrangement.