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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?

fennel
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1 Answers1

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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 $$f(\alpha x + (1-\alpha) y) + f(\alpha y + (1-\alpha) x) \ge f(x)+f(y)$$