Showing that if $u$ and $v$ are vectors in 2d-space or 3d-space, $∥u+v∥≤∥u∥+∥v∥$

Question

Use a theorem from plane geometry to show that if $u$ and $v$ are vectors in 2d-space or 3d-space, then $∥u+v∥≤∥u∥+∥v∥$.

I am a little stuck with this question, any hints would be greatly appreciated.

Answer 1

converting the vectors in complex numbers: u=e^(i(theta1)) v=e^(i(theta2)) convert them in the cosine and sine ratios and use the range for the cosine function.

Answer 2

I don't know if I can help you because I don't know what is meant by "Theorem from plane geometry", but you know for 2D it holds $||u||=\sqrt{u_1^2+u_2^2}$ because of the Pythagorean theorem.

If you write it down you get

$||u+v||^2=(u_1+v_1)^2+(u_2+v_2)^2\leq u_1^2+v_1^2+u_2^2+v_2^2=u_1^2+u_2^2+v_1^2+v_2^2=||u||^2+||v||^2$.

The $\leq$ holds because of the Binomial theorem.

It's the same calc. for 3D.