# Triangle Inequality on a different normed space

Let $x=(x_1,x_2)$, the norm is given by $[x]=\sqrt{x_1^2+x_1x_2+x_2^2}$

I need to show the triangle inequality holds.

So $y=(y_1,y_2)$ and from $[x+y]\le[x]+[y]$ I got $$4x_1^2y_1^2-6x_1x_2y_1y_2+4x_2^2y_2^2-x_1^2x_2^2-y_1^2y_2^2\ (\text{this must be bigger than }0)$$ after the long-time boring calculation. Then I got stuck! :(

Is there another way to show the triangle inequality or can we get the inequality from that equation?

-
thanks for the edit! I can't type those fancy letters... – Math-Nerd Apr 14 '14 at 10:58
I also changed $y=(y_1+y_2)$ to $y=(y_1,y_2)$, that was a typo, right? – user2345215 Apr 14 '14 at 11:01

Note that $$[x]=\sqrt{(x_1+{\textstyle\frac{1}{2}}x_2)^2+({\textstyle\frac{\sqrt3}{2}}x_2)^2}\ .$$ If you are permitted to use in your proof the triangle inequality for the standard Euclidean norm, you have $$\sqrt{(u_1+v_1)^2+(u_2+v_2)^2}\le\sqrt{u_1^2+u_2^2}+\sqrt{v_1^2+v_2^2}$$ for any real $u_1,u_2,v_1,v_2$. If you now substitute $$u_1=x_1+{\textstyle\frac{1}{2}}x_2\,,\quad u_2={\textstyle\frac{\sqrt3}{2}}x_2\,,\quad v_1=y_1+{\textstyle\frac{1}{2}}y_2\,,\quad v_2={\textstyle\frac{\sqrt3}{2}}y_2$$ you will get what you want.