Proof a function is convex if and only if it meets this condition?

Question

How to proof that a function f:R->R is convex if and only if dom(f) is convex and for each a, b, c in it's domain that are $a<b<c$, we have:

Determinant of matrix: $$ \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ f(a) & f(b) &f(c) \end{vmatrix}\ge 0. $$

Determinant is:

$$ bf(c)-cf(b)+cf(a)-af(c)+af(b)‐bf(a) >= 0$$

Then:

$$ f(a)(c-b) + f(b)(a-c) + f(c)(b-a) >=0$$

Then according to the a<b<c, we can say:

$$ f(a)(c-b) + f(c)(b-a) >= f(b)(c-a)$$ [edited]

So I went until here but I don't know how to connect this to the Jensen's inequality to prove that f is convex.

Answer 1

You need almost no more steps to go beyond there. The definition of convexity of a function includes the following: $$ f(\theta x+(1-\theta)y)\le \theta f(x)+(1-\theta)f(y) $$ Now, try to rearrange the last inequality you obtained with $x=a$, $y=c$ and a proper choice of $\theta$.