Strictly convexity of a function and global minima How to prove that a function $\mathbb{ f:R\to R}$ is strictly convex, then a critical point is a global minimum using Taylor expansion at the critical point?

Let us first show that $f$ can have only one critical point. By the first order condition for differentiable convex functions we know that for every $v,u$ we have 
$$f(u)>f(v)+f'(v)(u-v)$$.
If $x$ is a critical point then $f'(x)=0$ and thus every critical points is a strict minimum. Now suppose that there are two different local minima $x,y$. For $t \in (0,1)$ we then have 
$$f(tx+(1-t)y) < tf(x)+(1-t)f(y)$$
a contradiction to the fact that the minima are strict. Hence there is only one critical point which is a global minimum.
