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Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function. Find all functions which satisfy: $$f(x+y) - xy\ge f(x) +f(y)$$

And

$$ f(x) \ge 1-\cos(x) \quad \text{for any x,y real numbers}$$

I found out $f(0)=0$ and $f(x) \geq \frac{x^2}2$ but I got stuck.

Maximilian Janisch
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Jack
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    Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. Also, here's a MathJax tutorial – saulspatz Dec 13 '19 at 15:47
  • Thanks for editing – Jack Dec 13 '19 at 16:04
  • How do you get $f(x)\geq \frac{x^2}{2}$? I can see that $f(2x)\geq x^2$, but that's not the same thing, of course. – saulspatz Dec 13 '19 at 16:27
  • I ll post the method soon – Jack Dec 13 '19 at 16:44
  • I 've seen what s happening for f(x+y+z) and I generalised it by induction getting f(x/n+x/n+... +x/n) >=nf(x/n) +n(n-1)/2*x^2 /n^2 and passing to limit I got f(x) >=x^2 /2 because f(0)=0 – Jack Dec 13 '19 at 17:32
  • Overall, that is nicely thought out, but there is a problem. You know that $f\left(\frac xn\right) \to 0$ as $n \to \infty$. But that does not imply that $nf\left(\frac xn\right) \to 0$. But the inequality is still provable By the other restriction, $f\left(\frac xn\right) \ge 0$. So for all $n$, $$f(x) \ge \frac{x^2}2 - \frac{x^2}{2n}+ nf\left(\frac xn\right) \ge \frac{x^2}2- \frac{x^2}{2n}$$Letting $n \to 0$ now proves the $x^2/2$ inequality. – Paul Sinclair Dec 14 '19 at 01:15
  • Thank you! I ve also seen that f(x) +f(-x) <=x^2.and now I can prove that the function is x^2 /2 because f(x) +f(-x) >=x^2 /2+x^2/2-x^2 – Jack Dec 14 '19 at 06:53
  • @Jack - I thought of the same thing over breakfast this morning and was about to point it out. Glad you came up with it yourself first. – Paul Sinclair Dec 14 '19 at 13:50

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