Prove that if $X=[0,1]$ and $T:X \to X$ is defined as below then $d(Tx, Ty) \le \alpha (d(x, Tx) + d(y,Ty))$.

Question

$Tx = x/4$ if $x \in [0, 1/2)$ and $Tx =x/5$ if $x \in [1/2, 1]$ and $\alpha$ is a constant such that $0 \le \alpha < 1/2.$

What I have tried so far: if $x,y \in [0, 1/2)$ then

$d(Tx, Ty) = 1/4 d(x,y) \le 1/4 (d(x,Tx) + d(y,Ty) + d(Tx, Ty))$

This implies $d(Tx, Ty) \le 1/3 (d(x, Tx) + d(y, Ty))$ with $\alpha = 1/3$ And similarly if $x, y \in [1/2, 1]$ then $d(Tx, Ty) \le 1/4 (d(x, Tx) + d(y, Ty))$

I'm stuck with proving a similar result for when $x \in [0, 1/2), y \in [1/2, 1]$.