Given a polyhedron $P \subseteq \Bbb R^n$, a point $x \in P$ is called a vertex if there exists a vector $c$ such that $c · x > c · y$ for all $y \neq x \in P$.
A point $x \in P$ is called an extreme point if it is not the convex combination of two other points in $P$. That is, the equation $x = \lambda y + (1 − \lambda)z$ has no solution satisfying $y, z \neq x$ and $λ ∈ [0, 1]$.
We want to show that if $x$ is a vertex then $x$ is an extreme point. To do that I wanted to show the converse : if $x$ is not an extreme point then it is not a vertex :
If $x$ is not an extreme point then it is a convex combination of two other different points $y$ and $z$ : $x = \lambda y + (1 − \lambda)z$. We want to show that $\forall c \in \Bbb R^n$, $\exists w \in \Bbb R^n$ such that $c · x \leq c · w$
$c · x = c · (\lambda y + (1 − \lambda)z)=\lambda (c ·y)+(1-\lambda)(c·z)$
If $(c ·y)=0$ then $c · x = (1-\lambda)(c·z)=(c·(1-\lambda)z)$ so we can take $w=(1-\lambda)z$
If $(c ·y)<0$ then $c · x < (1-\lambda)(c·z)=(c·(1-\lambda)z)$ so we can take $w=(1-\lambda)z$
But I can't deal with the case $(c ·y)>0$, I feel that maybe $(1-\lambda)(c·z)$ will be negative but I struggle to show it.
I would separate the cases $\lambda\in]0,1[$ and $\lambda\in \{0,1\}$ instead, for instance:
Let $x$ be a vertex, and suppose that $x=\lambda y+(1-\lambda)z$ for some $\lambda\in]0,1[$, $y,z\in P\setminus\{x\}$, then \begin{align*} c\cdot x&=\lambda c\cdot y+(1-\lambda) c\cdot z\\ &> \lambda c\cdot x+(1-\lambda)c\cdot x\\ &= c\cdot x \end{align*} This is a contradiction and therefore there is no $\lambda\in ]0,1[$, $y,z\in P\setminus \{x\}$ such that $x=\lambda y+(1-\lambda)z$. Since $y,z\neq x$ then it is not possible to have $\lambda\in\{0,1\}$ and so there is no way to write $x=\lambda y+(1-\lambda)z$ with $y,z\neq x$ and $\lambda\in[0,1]$.