if $f'' >0$ prove that : $f(x+2)-f(x) \le f(x+5)-f(x+3)$ Let $f: R \to R$, differentiable twice such that $f'' > 0$ 
Prove that for every $ x>0$ exists:
$f(x+2)-f(x) \le f(x+5)-f(x+3)$
Any hints/suggestions? I got this problem at class and I couldn't figure out where to start.
 A: Suppose that $f(x+2) - f(x) > f(x+5) - f(x+3)$. By the mean value theorem there is $x_1 \in (x, x+2)$ such that $f'(x_1) = \frac{f(x+2) - f(x)}{2}$ and $x_2 \in (x+3, x+5)$ such that $f'(x_2) = \frac{f(x+5) - f(x+3)}{2}$. So $f'(x_1) > f'(x_2)$, which is a contradiction since $f'' > 0$ ($f'$ should be increasing).
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\dsc{\bracks{\fermi\pars{x + 2} - \fermi\pars{x}}
-\bracks{\fermi\pars{x + 5} - \fermi\pars{x + 3}}}
=\int_{x}^{x + 2}\fermi'\pars{t}\,\dd t
-\int_{x + 3}^{x + 5}\fermi'\pars{t}\,\dd t
\\[5mm]&=\int_{x}^{x + 2}\bracks{\fermi'\pars{t} - \fermi'\pars{t + 3}}
=-\int_{x}^{x + 2}\int_{t}^{t + 3}\fermi''\pars{s}\,\dd s\,\,\, \dsc{< 0}
\\[1cm]&\imp\quad\color{#66f}{\large%
\fermi\pars{x + 2} - \fermi\pars{x}
<\fermi\pars{x + 5} - \fermi\pars{x + 3}}
\end{align}
A: For a convex differentiable function $f$, the "incremental ratio"
$$\Delta_{x,y}=\frac{f(x)-f(y)}{x-y}, \quad \Delta_{x,x}=f'(x) $$
is a non-decreasing function with respect to both its variables, so:
$$\Delta_{x+2,x}\leq\Delta_{x+2,x+2}\leq\Delta_{x+3,x+3}\leq\Delta_{x+5,x+3}.$$
