curvature of the boundary of a convex set is positive Let's consider $J\subset \mathbb R^2$ such that J is convex and such that it's boundary it's a curve $\gamma$. Let's suppose that $\gamma$ is anti-clockwise oriented, let's consider it signed curvature $k_s$. I want to prove the intuitive following fact:
$$
\int\limits_\alpha  {k_s } \left( s \right)ds \geqslant 0
$$
For every sub-curve $\alpha \subset \gamma $.
And then prove that $k_s(s) \ge 0$
I have no idea how to attack this problem, intuitively I can see the result.
 A: This is a more formal version of Brian Rushton's answer. Suppose there is a point of negative curvature. Choose $xy$ coordinates so that this point is the origin $(0,0)$, the tangent direction is $x$-axis, and the $y$-axis points inside the convex set. Let $y=f(x)$ be the equation of a part of curve near $(0,0)$. (Implicit function theorem says you can solve for $y$ in terms of $x$.)
The curvature at $(0,0)$ is $f''(0)$, according to equation (14) here. Since $f''(0)<0$ and $f'(0)=0$,  it follows that $f(x)<0$ for $0<|x|<\delta$, if $\delta$ is sufficiently small. This contradicts the  convexity of the set: e.g., the line segment from $(x,f(x))$ to $(-x,f(-x))$ crosses the negative half of the $y$-axis.
A: If the curvature is negative, there must be a point with negative curvature. As you zoom up to that point, it looks more and more like the complement of a circle, which means that there are two points which are not connected by a straight line in the set.
A: If $s$ is arc-length, $T(s)$ is the unit tangent vector and $N(s)$ the counterclockwise unit normal, $\dfrac{d}{ds} T(s) = k(s) N(s)$.  It's convenient to consider the plane as the complex plane, so $T(s) = e^{i\theta(s)}$ and $N(s) = i e^{i \theta(s)}$.  Then we have
$\dfrac{d\theta}{ds} = k(s)$.  Now you want to show that $\theta(s)$ is nondecreasing...
