I am required to show that if the plane $\mathbb{R}^2$ is endowed with the metric $ds=\frac{|dz|}{1+|z|^2}$, then the shortest length path between the origin and a point $a \in \mathbb{R}$ is the straight line path between them. I'm getting stuck in this one.
Attempt:
Consider the straight line path from $0$ to $a$. Parametrise it as follows: $$ \gamma : [0,a] \to \mathbb{C} \quad \gamma(t)=t $$ Then we have \begin{align*} l(\gamma)&=\int_{0}^{a}\frac{|\gamma'(t)|}{1+|\gamma(t)|^2}dt\\ &=\int_{0}^{a}\frac{dt}{1+t^2}\\ &=\arctan(a) \end{align*}
Now consider an arbitrary path $\mu$ from $0$ to $a$ parametrised as follows: $$ \mu(t)=(x(t),y(t)), \quad 0\leq t\leq 1 $$ Since $y$ is continuous and $[0,1]$ is compact, there exists $M>0$ such that $|y(t)|\leq M$ for all $t\in[0,1]$. Then we have
\begin{align*} l(\mu)&=\int_{0}^{1}\frac{|\mu'(t)|}{1+|\mu(t)|^2}dt\\ &=\int_{0}^{1}\frac{\sqrt{x'(t)^2+y'(t)^2}}{1+x(t)^2+y(t)^2}dt\\ &\geq \int_{0}^{1}\frac{x'(t)}{1+M^2+x(t)^2}dt\\ &=\int_{0}^{a}\frac{du}{1+M^2+u^2}\\ &=\frac{1}{\sqrt{M^2+1}}\arctan\left(\frac{a}{\sqrt{M^2+1}}\right) \end{align*}
I cannot see how to conclude $l(\mu)\geq l(\gamma)$ from here. In fact, I think it cannot be concluded from here and I must be getting the estimations wrong. Cannot figure out what exactly though. Any hints would be appreciated!
HINT: Write this in polar coordinates, not cartesian.