Minkowski inequality proof without using Holder's inequality We are told to proof the Minkowski inequality specialized to sequences:
$$
\left(\sum_{i=1}^n|a_i+b_i|^p\right)^{1/p}\leq
\left(\sum_{i=1}^n|a_i|^p\right)^{1/p}
+\left(\sum_{i=1}^n|b_i|^p\right)^{1/p}
$$
when $p\geq1$ 
Well, the problem is that all the proofs I have found "use" the Holder's inequality to proof that, but we haven't seen it in class yet so I can't "use" it to proof the Minkowski inequality.
The proof must somehow "elementary", with things we have already learnt in class. "Cauchy-Schwartz inequality" and the "Triangle inequality"are two of the inequalities we have learnt, for example.
Could someone help me? (and the rest of the students in my class, who are in the same situation).
Thanks in advance!
 A: Denote
$$S := \sum_{i=1}^n |a_i|^p, \quad \mathrm{and}\quad T := \sum_{i=1}^n |b_i|^p.$$
If $S = 0$ or $T = 0$, the desired inequality is true.
In the following, assume that $S, T > 0$.
Let
$$t_0 = \frac{S^{1/p}}{S^{1/p} + T^{1/p}}.$$
Then $0 < t_0 < 1$. We have
\begin{align*}
 (S^{1/p} + T^{1/p})^p
 &= \frac{S^{1/p} + T^{1/p}}{(S^{1/p} + T^{1/p})^{1 - p}}\\
 &= \frac{S^{1/p}}{(S^{1/p} + T^{1/p})^{1 - p}} + \frac{ T^{1/p}}{(S^{1/p} + T^{1/p})^{1 - p}}\\
 &= t_0^{1-p} S
 + (1 - t_0)^{1-p}T\\
 &= \sum_{i=1}^n (t_0^{1-p}|a_i|^p + (1 - t_0)^{1-p}|b_i|^p)\\
 &= \sum_{i=1}^n \left[t_0\left(\frac{|a_i|}{t_0}\right)^p + (1 - t_0)\left(\frac{|b_i|}{1 - t_0}\right)^p\right]\\\
 &\ge \sum_{i=1}^n \left(t_0 \cdot \frac{|a_i|}{t_0} + (1 - t_0)\cdot \frac{|b_i|}{1 - t_0} \right)^p \tag{1}\\
 &= \sum_{i=1}^n (|a_i| + |b_i|)^p\\
 &\ge \sum_{i=1}^n |a_i + b_i|^p
\end{align*}
where in (1) we have used Jensen's inequality
for convex function $g(u) = u^p$ on $u\ge 0$.
We are done.

If Jensen inequality is not allowed, one may prove the following lemma.
Lemma. Let $1 \le p < \infty$ and $u, v \ge 0$. Then, for all $t \in (0, 1)$,
$$(u + v)^p \le t^{1 - p} u^p + (1 - t)^{1 - p}v^p.$$
(Hint: The nontrivial case is $p > 1$ and $u, v > 0$. Let $f(t) := t^{1 - p} u^p + (1 - t)^{1 - p}v^p$. Then $f'(t) = (1 - p)[t^{-p}u^p - (1 - t)^{-p}v^p]$. Then $f(t)$ has the global minimum at $t = u/(u+v)$.)
Reference.
[1] Heinz König, “A simple proof of the Minkowski inequality,” 1990.
A: For all $1\leq i\leq p$ ,$|a_i+b_i|\leq|a_i|+|b_i|$. Then we have that $$|a_i+b_i|^p \leq \left(|a_i|+|b_i|\right)^p.$$ Hence, $\displaystyle\sum_{i=1}^{p} |a_i+b_i|^p\leq\sum_{i=1}^{p}\left(|a_i|+|b_i|\right)^p$. Therefore, $$\left(\sum_{i=1}^{p} |a_i+b_i|^{p}\right)^{1/p}\leq\left(\sum_{i=1}^{p}|a_i|^p+\sum_{i=1}^{p}|b_i|^p\right)^{1/p}\leq\left(\sum_{i=1}^{p}|a_i|^p\right)^{1/p}+\left(\sum_{i=1}^{p}|b_i|^p\right)^{1/p}.\quad\blacksquare$$
