If $f'' \ge 0$ in interval $[0, 2]$, prove that $$\int_0^2 f(x)dx \ge 2f(1)$$

The question is graphically trivial I think, but not in mathematically.

I wanted to use the fact that there $\exists c$ s.t. $$\int_0^2f(x)dx = 2f(c)$$ and the Jensen's Thm, which is $$f(tx_1+(1-t)x_2)\le tf(x_1)+(1-t)f(x_2)$$

I tried to find the characteristics that $c$ can have, but it was still hard for me to get an idea. Could you please give some key points to the problem? Thanks.


By a function form of Jensen's inequality, it holds that $\frac{1}{2} \int_0^2f(x)dx \geq f(\frac{1}{2}\int_0^2xdx) = f(1)$. (See https://en.wikipedia.org/wiki/Jensen%27s_inequality)

  • $\begingroup$ +1. Nice proof. $\endgroup$ – Kavi Rama Murthy Feb 9 at 5:16
  • $\begingroup$ I think I got an idea. Thanks:D $\endgroup$ – ToBY Feb 9 at 5:18

$\int_0^2 f(x)dx$ =

$lim_{h\rightarrow 0}\ h[f(0)+f(0+h) + ........f(0 + (n-1)h)]\ =\ I$ where $nh=2$.

By jenson inquality-

$\frac{I}{\sum_{r=0}^{n-1}h}\ =\ \frac{I}{2}\ \ge\ $

$f(lim_{h\rightarrow 0}\ h.0 + h(0+h) + h(0+2h)...... h(0+(n-1)h)$ (where nh=2) =

$f(\int_0^2 xdx)$ = $f(1)$.


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