Prove that $$\displaystyle{(|x_1+y_1|^p + |x_2+y_2|^p +\dots +|x_n+y_n|^p)^{\frac{1}{p}}\leq (|x_1|^p + |x_2|^p +\dots +|x_n|^p)^{\frac{1}{p}}+(|y_1|^p + |y_2|^p +\dots +|y_n|^p)^{\frac{1}{p}}}$$ for all $p\in\Bbb{N}$.
I tried methods like taking the $p$th power on both sides and differentiating with respect to $x_1$, taking the rest of the variables as constants. I didn't get much farther.
A helpful hint would be great, and would suffice. I am not looking for a solution.
Thanks in advance!
This is known as Minkowski's Inequality. The easiest proof is using duality as is shown in this answer. The only tool needed there is Hölder's Inequality.
Hint
The expression in question is the $p$-Norm of vectors $x,y \in \mathbb{R}^n$ and translates to
$$\lVert x+y \rVert_p \leq \lVert x\rVert_p + \lVert y \rVert_p$$
Prove using the Cauchy-Schwarz inequality after taking the $p$-th power elementwise.
Hint: If $p\geq 1$, then the function $f:[0,\infty)\rightarrow[0,\infty),x\to x^p$ is convex. That is, for any $x,y\in[0,\infty)$ and $0\leq\theta\leq1$,
$$f(\theta x+(1-\theta)y)\leq \theta f(x)+(1-\theta)f(y).$$
