Let $f$ be a concave function. Then, by definition, for any $\alpha \in [0,1]$ \begin{equation} f(\alpha x + (1-\alpha) y) \geq \alpha f(x) + (1-\alpha) f(y) \end{equation}

Is there a way to prove that \begin{equation} f(x) + f(y) \leq f(\alpha x + (1-\alpha) y) + f(\alpha y + (1-\alpha) x) \end{equation}

by using the upper definition?

  • $\begingroup$ You titled this "convex functions" but then talk about "concave functions". Which is it? $\endgroup$
    – user247327
    Aug 31, 2020 at 13:13
  • $\begingroup$ Sorry for the confusion. I'm fixing the title $\endgroup$
    – fennel
    Aug 31, 2020 at 13:26

1 Answer 1


You have $$\begin{cases} f(\alpha x + (1-\alpha) y) &\ge \alpha f(x) + (1-\alpha) f(y)\\ f((1-\alpha) x + \alpha y) &\ge (1-\alpha) f(x) + \alpha f(y) \end{cases}$$

By adding those two inequalities you get the desired result $$f(\alpha x + (1-\alpha) y) + f(\alpha y + (1-\alpha) x) \ge f(x)+f(y)$$


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