:500px Let $\mathcal T$ be the tangent line to $f$ at some point $\left({c, f \left({c}\right)}\right)$, $c \in \left({a \,.\,.\, b}\right)$. From the Gradient-Intercept Form of a line, given any point $\left({x_1, y_1}\right)$ and the gradient $m$, the equation of such a line is: {{begin-eqn}} {{eqn | l = y - y_1       | r = m \left({x - x_1}\right)       | c =  }} {{eqn | ll=\implies       | l = y       | r = m \left({x - x_1}\right) + y_1       | c =  }} {{end-eqn}} For $\mathcal T$: * $y = \mathcal T \left({x}\right)$ * $y_1 = f \left({c}\right)$ * $m = f' \left({c}\right)$ * $x = x$ * $x_1 = c$ so :$\mathcal T \left({x}\right) = f' \left({c}\right) \left({x - c}\right) + f \left({c}\right)$ Consider the graph of $f$ to the right of $\left({c, f \left({c}\right)}\right)$, that is, any $x$ in $\left({c \,.\,.\, b}\right)$.  Let $d$ be the directed vertical distance from  $\mathcal T$ to the graph of $f$. That is, if $f$ is above $\mathcal T$ then $d > 0$. If $f$ is below $\mathcal T$, then $d < 0$.  (From the diagram, it is apparent that $\mathcal T$ is below $f$, but we shall prove it analytically.) $d$ can be evaluated by: {{begin-eqn}} {{eqn | l = d       | r = f \left({x}\right) - \mathcal T \left({x}\right)       | c =  }} {{eqn | r = f \left({x}\right) - f' \left({c}\right)\left({x - c}\right) - f \left({c}\right)       | c =  }} {{eqn | r = f \left({x}\right) - f \left({c}\right) - f' \left({c}\right) \left({x - c}\right)       | c =  }} {{end-eqn}} By the Mean Value Theorem, there exists some constant $k$ in $\left({c \,.\,.\, b}\right)$ such that: {{begin-eqn}} {{eqn | l = f' \left({k}\right)       | r = \frac {f \left({x}\right) - f \left({c}\right)} {x - c}       | c =  }} {{eqn | ll=\implies       | l = f' \left({k}\right) \left({x - c}\right)       | r = f \left({x}\right) - f \left({c}\right)       | c =  }} {{end-eqn}} Substitute this into the formula for $d$: {{begin-eqn}} {{eqn | l = d       | r = f' \left({k}\right) \left({x - c}\right) - f' \left({c}\right) \left({x - c}\right)       | c =  }} {{eqn | r = \left({f' \left({k}\right) - f' \left({c}\right)}\right) \left({x - c}\right)       | c =  }} {{end-eqn}} Recall that $x$ lies in the interval $\left({c \,.\,.\, b}\right)$. So $x > c$, and the quantity $x - c$ is (strictly) positive. $k$ is also in the interval $\left({c \,.\,.\, b}\right)$ and so $k > c$. By construction, $f$ is convex.  By the definition of convex: :$k > c \implies f' \left({k}\right) > f' \left({c}\right)$ which means that: :$\left({f' \left({k}\right) - f' \left({c}\right)}\right) > 0$ Then $d$ is the product of two (strictly) positive quantities and is itself (strictly) positive.  Similarly, consider the graph of $f$ to the left of $\left({c, f \left({c}\right)}\right)$, that is, any $x$ in $\left({a \,.\,.\, c}\right)$. By the same process as above, we will have: :$d = \left({f' \left({k}\right) - f' \left({c}\right)}\right) \left({x - c}\right)$ This time, $x < c$ and the quantity $x - c$ is (strictly) negative. Further, $k < c$, and so by a similar argument as above: :$k < c \implies f' \left({k}\right) < f' \left({c}\right)$ and the quantity $\left({f' \left({k}\right) - f' \left({c}\right)}\right)$ is also (strictly) negative. Thus $d$ will be the product of two (strictly) negative quantities, and will again be (strictly) positive.  {{qed}} {{proofread}}