Convexity of a Function Implies Function Lies Above Any Tangent Line I’m trying to use the following definition to show the result listed below.
Definition of Convex Function: Let $f$ be a real-valued function on an open interval. Let $a,b$ be points in the domain with $a<b$. Function $f$ is convex if $$f(a)+\frac{f(b)-f(a)}{b-a}(x-a)\ge f(x)$$ for each $x\in(a,b)$.
Goal: The function $f$ (defined as in the above definition) is convex if and only if $f(x)$ does not have any points below any tangent line for each $x$ in the domain. 
Attempted Proof: Suppose $f$ is convex. Let $x_1,x_2$ be points in the domain with $x_1<x_2$. From the definition of convexity given above, for each $x\in(x_1,x_2)$, $$f(x_1)+\frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)\ge f(x)$$
Let $x_*$ be a point between $x_1$ and $x_2$. For now, let $x\in(x_*,x_2)$. The tangent line to this point is given by $$t(x)=f(x_*)+f’(x_*)(x-x_*)$$
The Mean Value Theorem guarantees the existence of $c\in(x_*,x_2)$ such that $$f’(c)(x-x_*)=f(x)-f(x_*)$$
Next, consider the difference between the function and tangent line; our goal is to show this is nonnegative: $$f(x)-t(x)=f(x)-(f(x_*)+f’(x_*)(x-x_*))=f(x)-f(x_*)-f’(x_*)(x-x_*)=f’(c)(x-x_*)-f’(x_*)(x-x_*)$$
So, we have $$f(x)-t(x)=(x-x_*)(f’(c)-f’(x_*))$$
This is where I’m having difficulties performing any meaningful manipulation. If this can be shown true, the result for the left portion of the function can be shown analogously. 
 A: Without assuming that the convex function is twice-differentiable it is helpful to use a fundamental property that follows directly from the definition of convexity.  For $x < y < z$ we have
$$\frac{f(z) - f(y)}{z-y} > \frac{f(z) - f(x)}{z-x} > \frac{f(y) - f(x)}{y-x}$$
A convex function on an open interval always has one-sided derivatives, so
$$\frac{f(z) - f(x)}{z-x} \geqslant \lim_{y \to x+}\frac{f(y) - f(x)}{y-x} =f'_R(x),$$
giving you half of what you need.
If $f$ is differentiable and, hence, the tangent line is well-defined everywhere, then $f'_R(x) = f'_L(x)$ and it is easy to extend the argument to show that
$$f(z) \geqslant f(x) + f'(x)(z-x)$$ 
A: Fix $x_0\in (a,b)$ so that $f(x)=f(x_0)+f'(x_0)(x-x_0)$ is the equation of the tangent line. Suppose $y\in (a,b)$ is another point.  If $y=x_0$ the result is true trivially. If not, wlog $y>x_0$ and consider
$f(y)-f(x_0)-f'(x_0)(y-x_0)$. We need to prove that this is non-negative. Now, the mean value theorem applies to show that there is a $x_0<c<y$ such that the above is equal to $(f'(c)-f'(x_0))(y-x_0).$ Let us show that this expression is non-negative, which means proving that $f'(x_0)\le f'(c):$
First, choose $x_0<d<c$ and prove, from the definition of convexity,the chord-slope lemma, which will give us the inequalities
$\frac{f(d)-f(x_0)}{d-x_0}\le\frac{f(c)-f(x_0)}{c-x_0}\le\frac{f(c)-f(d)}{c-d}.$
Letting $d\to c^-$ we see that $\frac{f(d)-f(x_0)}{d-x_0}\le f'(c^-)=f'(c)$ because $f$ is assumed dfferentiable at $c$. 
$f'(c)$ is now a fixed number, so we can let $d\to x_0^+$ in $\frac{f(d)-f(x_0)}{d-x_0}$, and the inequality persists. We thus find that $f'(x_0^+)=f'(x_0)\le f'(c),\ $ as desired. 
