# Equality in Minkowski's inequality proof(no integrals)

So what I'm looking for is a proof for when does the equality hold in Minkowski's inequality? I'm talking about this form of inequality: $$\left( \sum_{K=1}^n |x_k + y_k|^p \right)^{\frac{1}{p}} \leq \left( \sum_{k=1}^n |x_k|^p\right)^{\frac{1}{p}} + \left( \sum_{k=1}^n |y_k|^p\right)^{\frac{1}{p}}$$

Not the one using integrals.

I know when does the equality hold, I just can't figure out how to prove it. It holds when (X1, ..., Xn) = c*(Y1, ..., Yn), where c is a real constant (this is similar to Holder's inequality which I used to prove Minkowski's inequality). So can I just point out that it similarly holds in Minkowski's case, as I suppose it does? Thanks.

EDIT:

I proved that if (X1, ..., Xn) = c*(Y1, ..., Yn) , then the equality holds. Now, how to do in reverse - if equality holds, then (X1, ..., Xn) = c*(Y1, ..., Yn) ? Because these are equivalent.

• Hi. You will probably get a much better response if you give some idea of what you have already tried - people will be happy to help you understand, but not to just do your homework for you. Also, do you just want to know when there is an equality, or a proof of this. Do you have any idea at to when it might hold? Commented Feb 26, 2014 at 19:52
• I don't think I will be able to solve this in general quickly, but some brief points: Firstly, if $p=1$ then the equality holds if and only if $x_i$ and $y_i$ are positive for all $i$. Secondly, I don't see what the equality necessarily holds if $(x_1, \ldots, x_n) = c(y_1, \ldots, y_n)$. For example, what if $c=-1$. Then the left hand side is zero, but the right hand side is not... Commented Feb 26, 2014 at 22:00

The case where $$p=1$$, the Minkowski's Inequality is nothing but a bunch of triangle inequalities of real numbers, that is $$\left( \sum_{K=1}^n |x_k + y_k|^p \right)^{\frac{1}{p}} \le \left( \sum_{k=1}^n |x_k|^p\right)^{\frac{1}{p}} + \left( \sum_{k=1}^n |y_k|^p\right)^{\frac{1}{p}},$$ when $$p=1$$, is $$\sum_{K=1}^n |x_k + y_k| \le \sum_{k=1}^n |x_k| + \sum_{k=1}^n |y_k|.$$

Equality holds when $$x_ky_k\ge0$$, for each k, that is $$x_k$$ and $$y_k$$ are either both non-negative or non-positive. So, that gives $$x_k=c_ky_k$$, for some positive $$c_k$$, for eack $$k$$.

For the case $$p>1$$ I assume you used \begin{align*} \left( \sum_{k=1}^n |x_k + y_k|^p \right) & =\left( \sum_{k=1}^n |x_k + y_k|^{p-1}|x_k + y_k| \right) \\ & \le \sum_{k=1}^n |x_k + y_k|^{p-1}|x_k|+\sum_{k=1}^n |x_k + y_k|^{p-1}|y_k|, \end{align*} that is a triangle inequality, followed by Hölder's inequality: $$\sum_{k=1}^n |x_k + y_k|^{p-1}|x_k| \le \left(\sum_{k=1}^n|x_k|^{p}\right)^{\frac{1}{p}}\left(\sum_{k=1}^n|x_k+y_k|^{(p-1)q}\right)^{\frac{1}{q}}, \quad \text{where } \frac{1}{p}+\frac{1}{q}=1.$$ and similarly, $$\sum_{K=1}^n |x_k + y_k|^{p-1}|y_k| \le \left(\sum_{k=1}^n|y_k|^{p}\right)^{\frac{1}{p}}\left(\sum_{k=1}^n|x_k+y_k|^{(p-1)q}\right)^{\frac{1}{q}}.$$

and added the expressions to get the desired result.

So, to have equality in Minkowski's inequality we must have equality in both applications of Hölder's inequality and the triangle inequality, i.e., $$\, \exists \, \lambda_1, \lambda_2 \ge 0$$ s.t., \begin{align*} |x_k|^p =\lambda_1|x_k+y_k|^{(p-1)q}, \qquad |y_k|^p =\lambda_2|x_k+y_k|^{(p-1)q} \tag{1} \end{align*} and \begin{align*} |x_k+y_k| =|x_k|+|y_k|, \tag{2} \end{align*} for $$k = 1,2, \cdots, n$$.

WLOG we may assume $$\lambda_1,\lambda_2 > 0$$. Therefore, from $$(1)$$ we have $$|x_k|^p = \left(\frac{\lambda_1}{\lambda_2}\right)|y_k|^p$$. Again for equality to hold in $$(2)$$ $$x_k$$ and $$y_k$$ must have same sign. So, that gives us the equality condition to be $$x_k = cy_k$$, for $$k = 1,2, \cdots, n$$ with $$c = \left(\frac{\lambda_1}{\lambda_2}\right)^{1/p}$$.

• Also pertaining the paragraph mentioned in the comment above: why do you need the triangle inequality to be an equality, too? $|x_k|^p = \frac{\lambda_1}{\lambda_2} | y_k|^p$ follows from the two "Hölder equalities", right? Commented May 5, 2019 at 0:40
• @ViktorGlombik You are right. I didn't use the equality of triangle inequality $(2)$ the first time.
– r9m
Commented May 5, 2019 at 4:07
• Thanks for the clarification. Here's one thing I still don't understand: Why can we conclude that if the sums in the Hölder inequality must be equal, all summands must be equal? I.e. we could have $1 + 4 = 2 + 3$ but obviously $1 \neq 2$ und $4 \neq 3$. Surely, one direction is true $x_k = y_k$ for all $k$ $\implies \sum_{k } x_k = \sum_{k} y_k$ but I'm not sure about the "only if" part. Commented May 5, 2019 at 18:15
• +1 for explaining why $x_k$ and $y_k$ must have same signs! Commented Sep 1, 2019 at 13:41
• @Ramanujan I think it goes as follows : You know $a\leq a'$ and $b\leq b'$. Then the equality $a+b=a'+b'$ implies the separate equalities $a=a'$ and $b=b'$. Commented May 1, 2021 at 20:01