Is it true, and i case of it being true, how can I show that:
(1) $\lambda_1x_1 +\cdots+\lambda_kx_k = \max\{x_1,\cdots,x_k\}$
(2) $\sum_{i=1}^{k}\lambda_i = 1$ and $\lambda_i \in (0 ,1] $ for $i = 1,\cdots, k$
Implies that $x_1 = x_2 = \cdots = x_k$ ?
Let $M= \max \{x_k \;|\; k=1,2,\dots ,n\}$. Then: $$x_k \le M$$ $$\lambda_k x_k \le \lambda_k M$$ $$\sum_{k=1}^{n} \lambda_k x_k \le \sum_{k=1}^{n} \lambda_k M = M$$
For the last inequality to turn into an equality, all the previous inequalities must be equalities as well (since any one strict inequality would cause the last one to be strict, too). Therefore $x_k = M$ for all $k=1,2, \dots,n$.
Let $x_1 = max\{x_1, x_2, \cdots, x_k\}$.
Then we have that
$$ \lambda_1x_1 + \lambda_2x_2 + \cdots \lambda_kx_k = x_1 \\ \lambda_1 + \lambda_2 + \cdots + \lambda_k = 1 $$
$$ \lambda_1x_1 + \lambda_2x_2 + \cdots \lambda_kx_k = x_1 \ (1) \\ \lambda_1x_1 + \lambda_2x_1 + \cdots + \lambda_kx_1 = x_1 \ (2). $$ Take $(1) - (2)$ to get that
$$ \lambda_2 (x_2-x_1) + \lambda_3 (x_3 - x_1) + \cdots + \lambda_k (x_k - x_1) = 0. $$
which can be written as
$$ \sum_{i=2}^{k} \lambda_i (x_i - x_1) = 0. $$
Since $\lambda_i > 0$ and $x_i - x_1 \leq 0$ we have that $\lambda_i (x_i - x_1) \leq 0$. From where we conclude that $\lambda_i (x_i - x_1) = 0$.
You must mean $\lambda_i> 0,$ not $\lambda_i\ne 0,$ otherwise it's false.
Let $M=\max (x_1,..,x_n).$ For each $i$ we have $$(1)\quad \lambda_i x_i\leq \lambda_iM.$$ Therefore $$(2) \quad \sum_{i=1}^n\lambda_i x_i\leq \sum_{i=1}^n\lambda_iM.$$ But if $x_i<M$ for any $i,$ we have strict inequality in (1) for that $i, $ and therefore strict inequality in (2). But the hypothesis is that we have equality in (2). Therefore $x_i=M$ for every $i.$
