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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?

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

3 Answers 3

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

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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 $$f(a)+f(b) \equiv (\beta f(a) + (1-\beta)f(b)) + ((1- \beta) f(a) + \beta f(b)) \ge f(\beta a + (1-\beta)b) + f((1-\beta)a + \beta b) \equiv f(a+c) + f(b-c).$$

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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.

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