Proving $f(x) \geq f(a) + f'(a) (x-a)$ I have this problem I'm working on:
Problem: Suppose $f: I \subset \mathbb{R} \rightarrow \mathbb{R}$ is a function that is differentiable at least twice on the interval $I$. Suppose that $f''(x) \geq 0$ for all $x \in I$. 
Prove that for all $a \in I$ and all $x \in I$ we have that $$ f(x) \geq f(a) + f'(a) (x-a). $$ 
Attempt: I was trying to use Mean value theorem of Lagrange and Taylor to prove this. But we probably will need the continuity of $f''$ aswell. Let $a \in I$ be arbitrary, and consider the interval $[a, x]$. Then Lagrange says there exists a $c \in ]a, x[$ such that $$ \frac{ f(x) - f(a)}{ x-a} = f'(c). $$ Now, I was trying to expand $f(x)$ in Taylor polynomials, but I'm not sure what to do with the Lagrange remainder, and/or if I should include it at all.
Any help/advice is appreciated!
 A: Solution 1
If $f''(x)\geq 0$ for all $x$, then $f$ is convexe, and thus $f$ is over it's tangent everywhere, i.e. $$\forall a,x\in\mathbb R, f(x)\geq f(a)+f'(a)(x-a).$$
The claim follow.
Solution 2
Since $f''(x)\geq 0$ for all $x$, there is $\theta\in ]0,1[$ s.t.
$$f(x)=f(a)+f'(a)(x-a)+\underbrace{\frac{f''((1-\theta)x+\theta a)}{2!}(x-a)^2}_{\geq 0}\geq f(a)+f'(a)(x-a),$$
what also prove the claim. 
Solution 3
Since $f''(x)\geq 0$ for all $x$, the function $f'$ is increasing. In particular, you can show that for all $y<a<x$
$$\frac{f(a)-f(y)}{a-y}\leq \frac{f(x)-f(a)}{x-a}.$$
If you let first $y\to a^-$, you'll get one part of the result. Then, if you let $x\to a^+$, you'll get the complete result.
A: For fixed $a$, let $g(x)=f(x)-(f(a)+f'(a)(x-a))$.  Note that $g'(x)=f'(x)-f'(a)$ and $g''(x)=f''(x)\ge0$.  This implies $g$ has a global minimum at $x=a$, since $g'(a)=f'(a)-f'(a)=0$, and thus $g(x)\ge g(a)=f(a)-(f(a)+f'(a)(a-a))=0$ for all $x$.  This implies $f(x)\ge f(a)+f'(a)(x-a)$ for all $x$.
