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Let $f$ be a continuous function, with the initial condition: $f''>0$.

I need to prove that $f(2x)-f(x) < f(3x)-f(2x)$.

By $f''>0$ I can learn that $f'$ is monotonous increasing, and so I tried to use Lagrange or Taylor.

Thank you.

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    By some theorem (I'm not sure which one) and under some conditions (I'm not sure which conditions) $f(2x) = f(x) + x f'(y)$ for some $0\leq y \leq x$ and similarly $f(3x) = f(2x) + x f'(y)$ for some $2x \leq y \leq 3x$.2011-03-01
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    By the way, It's kind of odd to refer to the mean value theorem as Lagranges theorem, I had to use wikipedia to know what you mean. (Where I've seen it, it is used for Lagranges theorem in group theory, or Lagranges four-square theorem.)2011-03-01
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    @Myself: What is meant is probably the Lagrange form of the remainder in the Taylor expansion, which indeed is (a generalization of) the mean value theorem.2011-03-01
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    @Myself: it's a common name for the mean value theorem outside the US.2011-03-01
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    @Andy: I don't even live in the US, perhaps you mean "outside Frenchspeaking countries"? Maybe Italy counts too, I remember Lagrange was actually Italian. :-) Thanks for the clarification though.2011-03-01
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    @Myself: sorry for the misunderstanding ;) I checked your profile and since there was nothing on it I made an assumption, no harm intended :D. Maybe it's for "romance language"-speaking countries, I don't know. ;)2011-03-01

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You are given that $f''(x)>0$ everywhere. If this condition holds, then $f(x)$ is convex on its domain.

Any convex function will satisfy Jensen's Inequality. Using Jensen's inequality with two terms, you have

\begin{equation} f(2x)<\frac{f(x)+f(3x)}{2} \end{equation}

i.e.,

\begin{equation} f(2x) - f(x) < f(3x) -f(2x) \end{equation}

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    It's nice to recognise the special case of Jensens inequality (I totally hadn't noticed) but doesn't it feel a bit like 'cheating', considering the "calculus I" environment?2011-03-01
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    If you are willing to use that $f'' > 0 \implies$ convexity, then isn't the desired inequality the "other", known to be equivalent for smooth functions, characterisation of convexity? (In other words, Jensen's inequality with two terms applied to convex functions is trivially true.)2011-03-01
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    @Myself - I didn't think I was cheating here because the OP didn't say anything about using only "Calculus I" methods.2011-03-01
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Let $g(y) = f(2x+y)$. Then $$g(y) = g(0) + y g'(0) + \frac{y^2}{2} g''(\xi)$$ $$g(-y) = g(0) - y g'(0) + \frac{y^2}{2} g''(\eta)$$ Note that $g'' > 0$. Hence, adding the above two we get $$g(y)+g(-y) = 2g(0) + \frac{y^2}{2} g''(\xi) + \frac{y^2}{2} g''(\eta) > 2g(0)$$

Hence, we get $$f(2x+y) + f(2x-y) > 2f(2x)$$ Plug in $y=x$ and rearrange to get $$f(2x)-f(x) < f(3x)-f(2x)$$

(PS: This is nothing but the derivation of Jensen's in the case when $\lambda = \frac{1}{2}$)

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I'll try to expand my comment a bit:

First assume $x>0$.

Since $f$ is continuous everywhere, it is continuous on $[x,2x]$. By the mean value theorem, there is some $y_1\in [x,2x]$ such that $f(2x)-f(x) = xf'(y_1)$. Similarly, there is some $y_2\in [2x,3x]$ suc that $f(3x)-f(2x) = xf'(y_2)$.

Therefore it is sufficient to show that $xf'(y_2) > xf'(y_1)$, or (because $x>0)$ that $f'(y_2) > f'(y_1)$. We know that $y_2 \geq y_1$ and that $f'$ is strictly increasing so we'll just have to exclude the possibility that $y_1 = y_2$. But in that case $y_2 = y_1 = 2x$ which leads to a contradiction. ( $f(2x) = f(x) + \int_x^{2x} f'(t)dt < f(x) + \int_x^{2x} f'(2x) dt = f(x) + xf'(2x) = f(2x)$. Although there are probably neater ways to find a contradiction.)

Now assume $x<0$. Construct $g(x) = f(-x)$ and reason with $g(x)$; I haven't done it but it shouldn't be too hard.