If z=x+i*y prove that |x|+|y|=

We'll raise to square both
sides:

(|x|+|y|)^2=<[(square root
2)*|z|]^2

(|x| + |y|)^2 = x^2 + y^2 +
2|x|*|y|

[(square root 2)*|z|]^2 =
2*|z|^2

But |z| = sqrt[Re(z)^2 + Im(z)^2], where Re(z) = x and Im(z)
= y

 |z| = sqrt(x^2 + y^2)

We'll raise
to square:

|z|^2 = x^2 + y^2

x^2 + y^2
+ 2|x|*|y| =< 2(x^2 + y^2)

We'll subtract x^2 + y^2 +
2|x|*|y| both sides and we'll use symmetric property:

2(x^2 + y^2) -
(x^2 + y^2 + 2|x|*|y|) > = 0

We'll remove the brackets and
we'll get:

2x^2 + 2y^2 - x^2 - y^2 -  2|x|*|y| >=
0

We'll combine like terms:

x^2 + y^2
-  2|x|*|y| >= 0

(|x| - |y|)^2 >= 0
true, so

|x|+|y|=<[(square root
2)*|z|]