$f(2x)-f(x)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.
 A: 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}
A: 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 $\lambda = \frac{1}{2}$)
A: 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.
