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.
 A: 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 $c = \left(\frac{\lambda_1}{\lambda_2}\right)^{1/p}$. 
