tag removed - Can one "soup-up" the LAW OF THE MEAN in the following way?

The answer is yes. Suppose the contrary and rescale the picture so that $r=1$. We may assume that $P_1$ and $P_3$ are the endpoints of the graph. There must be points on the graph that are outside the circle - otherwise the curvature at $P_2$ is at least 1. WLOG assume that there are points below the circle. The lower half of the circle is the graph of a function $f_0$ defined on an interval of length 2. Let $q$ be a point where $f_0-f$ attains its maximum, then $f'(q)=f_0'(q)$. We may assume that $f'(q)=f_0'(q)ge 0$, otherwise reflect the picture through the $y$-axis.

I claim that $f(t)<f_0(t)$ for all $tge q$ such that both $f(t)$ and $f_0(t)$ are defined, contrary to the fact that $P_2$ lies on the circle. To prove this, consider arc-length parametrizations $gamma(t)=(x(t),y(t))$ and $gamma_0(t)=(x_0(t),y_0(t))$ of the two graphs, where $gamma(0)=(q,f(q))$ and $gamma_0(0)=(q,f_0(q))$. Then
$$
gamma'(0)=gamma_0'(0)=(cosalpha,sinalpha)
$$
where $0lealpha<pi/2$. Since the curvature of $gamma$ is less than 1, we have
$$
angle(gamma'(t),gamma'(0)) < t
$$
for all $t>0$. Therefore $x'(t)>cos(alpha+t)$ and $y'(t)<sin(alpha+t)$ for $0<t<pi/2-alpha$. And for $gamma_0$ these inequalities turn to equalities. The integration yields that $x(t)ge x_0(t)$ and $y(t)< y_0(t)$ for all $tin[0,pi/2-alpha]$. Since $f_0$ is increasing after $q$, these inequalities imply that $gamma(t)$ is below the half-circle (or has already left the domain where $f_0$ is defined).

Since $gamma_0(pi/2-alpha)$ is the rightmost point of the circle, the inequality $x(t)ge x_0(t)$ for $t=pi/2-alpha$ means that the $x$-coordinate of $gamma(pi/2-alpha)$ is already outside the domain of $f_0$ and one does not need to care about larger values of $t$.