Let $\mathcal{A}$ be the family of closed smooth curves in the right half of the complex plane $\mathbb{C}$ such that any curve in the family must enlose the point $z=1$ and tangent to the $y$-axis at the origin. Then we define the weighted length of curves in the family as
$$L(\gamma):= \int\_{\gamma} \frac{2}{1+|z|^2}d|z|,$$where $d|z|$ is the classical length element.


My question is that, is it true that $\inf\_{\gamma \in \mathcal{A}}L(\gamma) =\pi$? 


I have done some computations for some curves with explicit formulas. For example, if $\gamma$ is a unit circle centered at $z=1$, then $\gamma \in \mathcal{A}$ and $L(\gamma)=4\pi/\sqrt{5}$. Also, it seems that when $\gamma$ is more and more closed to the line segment $[0,1]$ with multiplicity 2, the weighted length $L(\gamma)$ is getting smaller and approaching to $\pi$. 


Any ideas or comments are really appreciated. Thank you very much for your time.



minimizing weighted length of closed curves
I just found out two proofs. One is analytic, and the other is geometric.


For the analytic proof, one can use polar coordinates and apply some elemetary inequality. But such proof has limited applications. so I'm presenting a geometric proof.


Note that the metric of the unit sphere is given by $g=\frac{4}{(1+|z|^2)^2}\delta$, where $z$ is the coordinate of point on sphere obtained using stereographic projection map from the north pole, and $\delta$ is the Euclidean metric. Hence $L(\gamma)=l\_g(\Pi^{-1}(\gamma))$, where $\Pi$ is the stereographic projection map and $l\_g$ is the length function on the unit sphere. For any $\gamma \in \mathcal{A}$, since $\gamma$ enloses the point $z=1$, $\Pi^{-1}(\gamma)$ must be a closed curve on sphere suth that it starts at the south pole, and it must goes to some point in the northern hemisphere and then goes back to the south pole. Therefore, $\Pi^{-1}(\gamma) \ge \pi$, and $"="$ holds if and only if $\gamma$ is a line segment connecting $z=0$ and $z=1$.


