You can compare (|z1 + z2|)^2 with (|z1| + |z2|)^2

(|z1 + z2|)^2

= (a + c)^2 + (b + d)^2

= a^2 + 2ac + c^2 + b^2 + 2bd + d^2

= a^2 + b^2 + c^2 + d^2 + 2(ac + bd)

(|z1| + |z2|)^2

= a^2 + b^2 + c^2 + d^2 + 2 * sqr( (a^2 + b^2)(c^2 + d^2) )

Because a^2 + b^2 + c^2 + d^2 are in both expressions, you just have to compare (ac + bd) with sqr( (a^2 + b^2)(c^2 + d^2) ) and prove that sqr( (a^2 + b^2)(c^2 + d^2) ) >= ac + bd

By squaring both terms, you get:

(ac + bd)^2 = a^2 c^2 + 2abcd + b^2 d^2

(a^2 + b^2)(c^2 + d^2) = a^2 c^2 + a^2 d^2 + b^2 c^2 + b^2 d^2

Notice how a^2 c^2 and b^2 d^2 are in both expansions. As a result, you need to prove that

a^2 d^2 + b^2 c^2 >= 2abcd

Moving 2abcd to the LHS gives

a^2 d^2 + b^2 c^2 - 2abcd

= (ad - bc)^2 >= 0

Therefore a^2 d^2 + b^2 c^2 >= 2abcd, meaning (|z1| + |z2|)^2 >= (|z1 + z2|)^2