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I've been given this question:

Let $f(x)$ be a twice differentiable function. Suppose that $f''(x) \gt 0$ For every $x \in [0, \infty)$.

Prove that for every $x \gt 0$:

$f(x+2) - f(x) \lt f(x+5) - f(x+3)$

What do I know here? We recently learned about the Mean-Value theorem. Can it actually help me here?

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    hint: since $f''$ is positive, $f'$ is increasing. LHS and RHS seems to be closely related to $f'$ – 2017-02-04
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    Another hint : The function f seems to be accelerating – 2017-02-04

3 Answers 3

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You can apply the mean value theorem on the intervals $[x, x+2]$ and $[x+3, x+5]$ to get that there is $c_1 \in [x,x+2]$ such that

$$ f'(c_1)((x+2)-x) = f(x+2)-f(x) \implies 2f'(c_1)=f(x+2)-f(x)$$

and similarly $c_2 \in [x+3,x+5]$ such that

$$ 2f'(c_2)=f(x+5)-f(x+3) $$

As mentioned in the comments on your question, $f''(x) > 0$ means that $f'$ is increasing, so

$$ a < b \implies f'(a) < f'(b) $$

Can you fill in the rest?

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    Good answer :) I wrote mine on my phone so I was unable to quickly fill in all of the MathJax details, and was just about done when I saw your post. Also, while I'm at it: Welcome to Math.SE! Thanks for taking your time to help out others in the community. If you have any questions about anything site-related feel free to message me (a comment below my post will do) or write up a question on Meta.Math.SE. – 2017-02-04
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Your function is positive, so it must be increasing, as must be the derivative $f'(x)$. If we look at $f(x+2) - f(x) \lt f(x+5) - f(x+3)$ we see that we can write this as $$f(x+2) - f(x) \lt f(v+2) - f(v)$$ We can now use the fact that $xx$, the RHS must be greater than the LHS.

Think of it this way: since the derivative is increasing, we know that the farther we go to the right the faster the function is growing. What we are looking at with the subtraction is the difference in heights between points separated by two, which is pretty small at first but gets larger and larger. If you want a visual, imagine $f'(x) = x^2$ and try drawing the heights.

Proof

First of all, note that $f''(x)>0 \implies f'(x)$ is increasing, and let $v=x+3$ as above

Note that we can use the Mean Value Theorem on both $x$ and $v$ $$2f'(c_v) = f(v+2)-f(v)$$ $$2f'(c_x) = f(x+2)-f(x)$$ We note that $$v = x+3 < c_v < v+2 = x+5 \implies 3 < c_v - x < 5$$ $$v = x < c_x < x+3 \implies 0 < c_x - x < 2$$ Now, the inequalities above show that $c_v-x > c_x-x \implies c_v > c_x$. We also know that, since $f'(x)$ is increasing, $a

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This question might be answered more easily by introducing another function $g(x)=f(x+2)-f(x)$. The MVT gives you that $\frac{f'(x+2)-f'(x)}{(x+2)-x}=f''(c)$ for some $c\in(x,x+2)$. Since $f''(c)>0$ on that interval, we can also conclude $g'(x)=f'(x+2)-f'(x)>0$ on that interval.

Applying the MVT once more, we find that $\frac{g(x+3)-g(x)}{(x+3)-x}=g'(c)$ for some $c\in(x,x+3)$. Since $g'(c)>0$, we also conclude $g(x+3)-g(x)>0$. Re-writing $g(x+3)>g(x)$ in terms of $f$ gives you your desired inequality.