Inequality for convex functions

Question

If $f$ is a convex function then, for all $a<b$ and $0\le c<b-a$, $$f(a)+f(b)\ge f(a+c)+f(b-c).$$ What is the shortest proof for the inequality?

Answer 1

Let $a<b$ and $0<c< b-a$, suppose first to all that $a+c\leq b-c$ a property of convex functions in one variable says that $$ \frac{f(b-c)-f(a)}{b-c-a}\leq\frac{f(b)-f(a+c)}{b-c-a} $$ then $$ f(b-c)+f(a+c)\leq f(b)+f(a) $$

If $b-c<a+c$ then $$ \frac{f(a+c)-f(a)}{a+c-a}\leq\frac{f(b)-f(b-c)}{b-b+c} $$

Answer 2

There seem to be the simplest proof: for some $\beta,\ 0\le \beta\le 1$, $$a+c=\beta a + (1-\beta)b,$$ $$b-c=(1-\beta)a + \beta b.$$ Then $$f(a)+f(b) \equiv (\beta f(a) + (1-\beta)f(b)) + ((1- \beta) f(a) + \beta f(b)) \ge f(\beta a + (1-\beta)b) + f((1-\beta)a + \beta b) \equiv f(a+c) + f(b-c).$$

Answer 3

By given inequalities

$$(b,a) \succ (b-c, a+c) \;\;\;\; \text{or} \;\;\;\; (b,a) \succ (a+c, b-c)$$

holds. Using Karamata's Majorization Inequality gives the desired inequality.