Condition for $x,y$ so $|x+y| = C(|x| + |y|)$

Question

I want to find conditions on $x,y$ so the equation $|x+y| = C(|x|+|y|)$ will hold for $0<C\leq 1$. For $C = 1$, we have equality if and only if $\exists c>0$ s.t. $x = cy$. I tried to follow a similar path on proving the result for $C=1$: If equality holds, then $$ x^2+y^2+2xy = C^2(x^2+y^2 + 2|x||y|) \Rightarrow \frac{1-C^2}{2}(x^2+y^2) = C^2|x||y| - xy $$ Assume $C\neq 1$. Then either $x=y=0$ or both are not zero. So assume both are not zero. Then the left hand side is positive non-zero, so the right hand side must also be positive non zero. This means that $C^2|x||y| > xy$. So the only thing I learn from this is that because the inequality is strict and $C<1$, $x,y$ must have opposite signs. But I can't seem to find a neat condition depending on the value of $C$.

Answer 1

Let $x,y\ge 0$ then we have$$x+y=C(x+y)$$for $C=1$ this holds for any $x,y\ge 0$ and for $C<1$ this holds only if $x=y=0$.

Let $x\ge0\quad ,\quad y<0$ therefore$$|x+y|=C(x-y)$$for $x>-y$ we have $$x+y=C(x-y)\to y=\dfrac{C-1}{C+1}x\to -x<y\le 0$$which is a contradiction. Now suppose $x\le -y$ therefore $$-x-y=C(x-y)\to y=\dfrac{C+1}{C-1}x\to-\infty<y<-x$$

Conclusion

For $C=1$ the only case is where $xy\ge 0$ and for $0<C<1$ we must have$$\text{either }x\ge 0\text{ and }-\infty<y<-x\\\text{or }y\ge 0\text{ and }-\infty<x<-y$$