If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|. If 'z' and 'w' are two complex numbers, prove that: |z+w|<_|z|+|w|.
My Attempt; 
If z is given as x + iy, and w = u + iv,
then |z| = √(x^2 + y^2), = r, say;
and |w| = √(u^2 + v^2), = p, say
Then we can also view z as r cis θ,
and w as p cis φ (i.e., p (cos φ + i sin φ) in
polar form; so that θ and φ can be
determined from x, y, u, and v -- to within
2π.)
Now if z + w = q, say,
consider the triangle formed by z, w, and
q.
The angle opposite q is π - |φ - θ| .
So we have |q|^2 = |z|^2 + |w|^2 + 2 |
Is this correct. Please somebody check it. And if there is any other method... then plaese do it.
z| |w| cos (θ + φ),
which is < |z + w|^2, since cos x

Perhaps writing the expression explicitly it can be clearer:
$$\begin{cases}z=a+bi\\{}\\w=x+iy\end{cases}\implies |z+w|=\sqrt{(a+x)^2+(b+y)^2}\stackrel?\le\sqrt{a^2+b^2}+\sqrt{x^2+y^2}$$
Well, now square both sides:
$$a^2+2ax+x^2+b^2+2by+y^2\stackrel?\le a^2+b^2+x^2+y^2+2\sqrt{(a^2+b^2)(x^2+y^2)}\iff$$
$$\iff2(ax+by)\stackrel?\le2\sqrt{(a^2+b^2)(x^2+y^2)}\iff$$
$$\iff a^2x^2+2abxy+b^2y^2\le a^2x^2+a^2y^2+b^2x^2+b^2y^2\iff$$
$$2abxy\le a^2y^2+b^2x^2\iff a^2y^2+b^2x^2-2abxy\ge0\iff$$
$$\iff (ay-bx)^2\ge0\;\;\;\color{green}\checkmark$$
