How to prove this sign of the derivative? Suppose that $u:[0,\delta]\rightarrow\mathbf{R}$, $u\in C^2((0,\delta))\cap C([0,\delta])$ such that $$u(0)=0,$$ $$u>0 \ \ in \  (0,\delta],$$$$u''>0$$ Then $u'>0$ in $(0,\delta)$.
 A: The condition $u''>0$ tells that the function is concave, that is the curve is facing upwards in the coordinate axes. 
Just like $y=x^2$. For this to happen while $u>0$ be positive, while $u(0)=0$, 
$u$ either has a local minima at $0$ or is merely increasing from $0$ towards the positive side of $x$-axis. 
So it's 1st derivative is positive when you are approaching the curve at $0$ from the positive side. 
Just imagine the curve with these conditions. 
A: Firstly, $u''>0$ in $(0,\delta)$ implies that $u'$ is strictly increasing in $(0,\delta)$. Hence we need only show that for any given $x \in (0,\delta)$, $u'(x)\geq 0$.
So let $x \in (0,\delta)$. By MVT, $\exists c\in(0,x)$ such that $u'(c)=\frac{u(x)-u(0)}{x-0}=\frac{u(x)}{x}>0$ as $u(0)=0$ and $u(x)>0$. As $u'$ is strictly increasing this implies $u'(x)>0$ as needed.
A: Suppose that there is no such $\epsilon$. Then on every interval of the form $(0,\epsilon)$, and in particular on the interval $(0,\delta)$, there is a number $x_0$ such that $u'(x_0)\leq0$. As noted, we can take $x_0\in(0,\delta)$. Since $u''>0$ on this interval, we find $u'(x)<0$ for all $x\in(0,x_0)$. Now apply the mean value theorem to $u$ at two points $x_1,x_2\in(0,x_0)$ to obtain a contradiction. 
