Let $f$ be strictly concave real function on $I$ according to definition 1. That is: :$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) > \alpha f \left({x}\right) + \beta f \left({y}\right)$ Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$. As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \left({1 - \alpha}\right) x_3$. Thus: :$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$ So: {{begin-eqn}} {{eqn | n = 1       | l = f \left({x_2}\right)       | o = >       | r = \frac {x_3 - x_2} {x_3 - x_1} f \left({x_1}\right) + \frac {x_2 - x_1} {x_3 - x_1} f \left({x_3}\right)       | c = substituting for $\alpha$ and $\beta$ }} {{eqn | ll= \iff       | l = \left({x_3 - x_1}\right) f \left({x_2}\right)       | o = >       | r = \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)       | c =  }} {{eqn | ll= \iff       | l = \left({x_3 - x_1}\right) f \left({x_2}\right)       | o = >       | r = \left({x_3 - x_1 - x_2 + x_1}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)       | c =  }} {{eqn | ll= \iff       | l = \left({x_3 - x_1}\right) f \left({x_2}\right) - \left({x_3 - x_1}\right) f \left({x_1}\right)       | o = >       | r = \left({x_2 - x_1}\right) f \left({x_3}\right) - \left({x_2 - x_1}\right) f \left({x_1}\right)       | c =  }} {{eqn | ll= \iff       | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}       | o = >       | r = \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}       | c =  }} {{end-eqn}} Again from $(1)$: {{begin-eqn}} {{eqn | n = 1       | l = f \left({x_2}\right)       | o = >       | r = \frac {x_3 - x_2} {x_3 - x_1} f \left({x_1}\right) + \frac {x_2 - x_1} {x_3 - x_1} f \left({x_3}\right)       | c =  }} {{eqn | ll= \iff       | l = \left({x_3 - x_1}\right) f \left({x_2}\right)       | o = >       | r = \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)       | c =  }} {{eqn | ll= \iff       | l = \left({x_3 - x_1}\right) f \left({x_2}\right)       | o = >       | r = \left({x_3 - x_1 - x_3 + x_2}\right) f \left({x_3}\right) + \left({x_3 - x_2}\right) f \left({x_1}\right)       | c =  }} {{eqn | ll= \iff       | l = \left({x_3 - x_2}\right) f \left({x_3}\right) - \left({x_3 - x_2}\right) f \left({x_1}\right)       | o = >       | r = \left({x_3 - x_1}\right) f \left({x_3}\right) - \left({x_3 - x_1}\right) f \left({x_2}\right)       | c =  }} {{eqn | ll= \iff       | l = \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}       | o = >       | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}       | c =  }} {{end-eqn}} Thus: :$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$ So: :$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$ demonstrating that $f$ is strictly concave on $I$ according to definition 3, and: :$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$ demonstrating that $f$ is strictly concave on $I$ according to definition 2. As each step is an equivalence, the argument reverses throughout. {{qed}} Category:Concave Real Functions ilxskpfcipiah0tt4dl464e0w4e2lrp