Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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: 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 $0\leq\theta\leq1$,
$$f(\theta x+(1-\theta)y)\leq \theta f(x)+(1-\theta)f(y).$$
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.
