Prove the inequality for strictly convex function 
Suppose that $g:\mathbb{R}^s \to \mathbb{R}$ is strictly convex function, then I need to show that there exists $\beta\in \mathbb{R}^s$ such that $$g(x)>g(y)+\sum \beta_i (x-y)_i$$ $\forall x\neq y$ 

I could prove this result when $g$ is differentiable enough, using the Taylor series , but I don't know how to do this for not differentiable $g$.
 A: Here is an approach that relies on the separation theorem.
Let $\operatorname{epi} g$ be the epigraph of $g$, note that it is a convex set. The point $(y, g(y))$ is not in the interior of $\operatorname{epi} g$ so there is a hyperplane separating $(y, g(y))$ and $\operatorname{epi} g$.
In particular, there is some $(h,h_0) \neq 0$ such that
$\langle (h,h_0), \gamma -(y, g(y)) \rangle \ge 0$ for all $\gamma \in \operatorname{epi} g$. Since $(y, g(y)+1) \in \operatorname{epi} g$, we see
that $h_0 \ge 0$. If $h_0 = 0$, then $\langle (h,0), \gamma -(y, g(y)) \rangle =  \langle h, x-y \rangle\ge 0$ for all $x$ hence $h=0$, a contradiction.
Hence $h_0 >0$, by dividing through we can assume $h_0 = 1$.
Since $(x,g(x)) \in \operatorname{epi} g$, we have $g(x)-g(y) \ge \langle -h, x-y \rangle$, and letting $\beta = -h$, we have
$g(x)-g(y) \ge \langle \beta , x-y \rangle$.
Note that this implies $g(y+t(x-y))-g(y) \ge t \langle \beta , x-y \rangle$
for $t \in [0,1]$.
All that remains is to show strictness. Suppose for some $x\neq y$ we have
$g(x)-g(y) = \langle \beta , x-y \rangle$. Then, for $t \in [0,1]$,
$t \langle \beta , x-y \rangle \le g(y+t(x-y))-g(y) \le t (g(x)-g(y)) = t \langle \beta , x-y \rangle$, and so
$g(y+t(x-y))-g(y) = t \langle \beta , x-y \rangle$, which contradicts
the strictness of $g$. Hence
$g(x)-g(y) > \langle \beta , x-y \rangle$ when $x \neq y$.
