21think geometrically
proving the RHS
write the given expression as |z-1| + |1+ z2| ≤ |z-1| + |1| + |z2| (equality can hold)
|z-1| + |1| + |z2| = 2 + |z-1|≤ 4 because |z-1|≤2 (say graphically by finding the max distance between z and (1,0) where z lies in a circle of unit radious or simply use modulus inequality)
so we get given expression less than equal to 4
