Let $x_1, x_2, x_3 \in \R$ such that $x_1 < x_2 < x_3$. Consider the expressions: :$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}$ :$\dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$ The following cases are investigated: :$(1): \quad x_1, x_2, x_3 < 0$: Then: {{begin-eqn}} {{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | r = \frac {-\left({x_2 - x_1}\right)} {x_2 - x_1}       | c = {{Defof|Absolute Value}} }} {{eqn | r = -1       | c =  }} {{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | r = \frac {-\left({x_3 - x_2}\right)} {x_3 - x_2}       | c = {{Defof|Absolute Value}} }} {{eqn | r = -1       | c =  }} {{eqn | ll= \implies       | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | o = \le       | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | c = {{Defof|Convex Real Function}} }} {{end-eqn}} :$(2): \quad x_1, x_2, x_3 > 0$: Then: {{begin-eqn}} {{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | r = \frac {x_2 - x_1} {x_2 - x_1}       | c = {{Defof|Absolute Value}} }} {{eqn | r = 1       | c =  }} {{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | r = \frac {x_3 - x_2} {x_3 - x_2}       | c = {{Defof|Absolute Value}} }} {{eqn | r = 1       | c =  }} {{eqn | ll= \implies       | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | o = \le       | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | c = {{Defof|Convex Real Function}} }} {{end-eqn}} :$(3): \quad x_1 < 0, x_2, x_3 > 0$: {{begin-eqn}} {{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | r = \frac {x_2 - \left({-x_1}\right)} {x_2 - x_1}       | c = {{Defof|Absolute Value}} }} {{eqn | r = \frac {\left({x_2 + x_1}\right) + \left({x_1 - x_1}\right)} {x_2 - x_1}        | c =  }} {{eqn | r = \frac {x_2 - x_1} {x_2 - x_1} + \frac {2 x_1} {x_2 - x_1}       | c =  }} {{eqn | r = 1 + \frac {2 x_1} {x_2 - x_1}       | c =  }} {{eqn | o = <       | r = 1       | c = as $2 x_1 < 0$ }} {{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | r = \frac {x_3 - x_2} {x_3 - x_2}       | c = {{Defof|Absolute Value}} }} {{eqn | r = 1       | c =  }} {{eqn | ll= \implies       | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | o = \le       | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | c = Definition of Convex Real Function }} {{end-eqn}} :$(4): \quad x_1, x_2 < 0, x_3 > 0$: {{begin-eqn}} {{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | r = \frac {-\left({x_2 - x_1}\right)} {x_2 - x_1}       | c = {{Defof|Absolute Value}} }} {{eqn | r = -1       | c =  }} {{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | r = \frac {x_3 - \left({-x_2}\right)} {x_3 - x_3}       | c = Definition of Absolute Value }} {{eqn | r = \frac {x_3 + x_2 + \left({x_3 - x_3}\right)} {x_3 - x_2}        | c =  }} {{eqn | r = \frac {-x_3 + x_2 + x_3 + x_3} {x_3 - x_2}        | c =  }} {{eqn | r = \frac {-\left({x_3 - x_2}\right)} {x_3 - x_2} + \frac {2 x_3} {x_3 - x_2}       | c =  }} {{eqn | r = -1 + \frac {2 x_3} {x_3 - x_2}       | c =  }} {{eqn | o = >       | r = -1       | c = as $2 x_3 > 0$ }} {{eqn | ll= \implies       | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | o = \le       | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | c = {{Defof|Convex Real Function}} }} {{end-eqn}} Thus for all cases, the condition for $f$ to be convex is fulfilled. Hence the result. {{qed}}