# Inequality for convex functions

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

• Any proof would have to copy with the counterexample $f(x)=x^2$, $a=0$, $b=98$, $c=99$, $d=100$. Apr 23, 2019 at 15:53
• @HagenvonEitzen thank you for the comment. I have corrected the inequality. Apr 23, 2019 at 16:02
• $c$ is positive or can be also negative? Apr 23, 2019 at 16:57
• @Jihlbert Thank you! Corrected. Apr 23, 2019 at 17:04

Let $$a and $$0, 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 $$b-c then $$\frac{f(a+c)-f(a)}{a+c-a}\leq\frac{f(b)-f(b-c)}{b-b+c}$$
There seem to be the simplest proof: for some $$\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).$$
$$(b,a) \succ (b-c, a+c) \;\;\;\; \text{or} \;\;\;\; (b,a) \succ (a+c, b-c)$$