# Prove an inequality.

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.

• This seems similar to this question.
– robjohn
Aug 6, 2013 at 18:55

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.

• It's funny, that is the exact thing I was trying to prove, which led me to this inequality. So I doubt that would be very helpful in my particular circumstance. Thanks!
– user67803
Aug 6, 2013 at 17:29
• I edited. The standard way utilizes the CS-inequality. Aug 6, 2013 at 17:30
• @AlexR I'm pretty sure you can only use the CS inequality here if $p=2$, the case in which the norm is "induced" by an inner product.
– jkn
Aug 6, 2013 at 17:41
• @jkn For other $p$, use Minkowski. The proof is ibid. BTW. So if anyone wants to C&P the answer from wikipedia, go ahead. Aug 6, 2013 at 17:45
• @AlexR, that is the inequality the OP is trying to prove.
– jkn
Aug 6, 2013 at 17:48