$\Phi$ a quadratic form, then $\sqrt{\Phi (x+y)} = \sqrt{\Phi (x)} + \sqrt{\Phi (y)}$ if $x= \lambda y for x,y, \lambda \in \mathbb{K}$

Question

I'm trying to prove or disprove the following statement:

$\Phi$ a positive quadratic form on $\mathbb{K}^n$, then $\sqrt{\Phi (x+y)} = \sqrt{\Phi (x)} + \sqrt{\Phi (y)}$ if $x= \lambda y$ for $x,y, \lambda \in \mathbb{K}$ and $\lambda \neq 0$

Since $\Phi$ is a quadratic form, it can be rewritten as:

$\Phi(x) = ax^2 + bx + c$

$\Phi (y) = dy^2+ey + f$

$\Phi(x+y) = g(x+y)^2 + h(x+y) + j$

Where $a,b,c,d,e,f,g,h,j$ are constants.

Am I allowed to rewrite the quadratic form $\Phi$ in such a way if I don't know the explicit function?

Answer 1

Hint.

If $x = \lambda y$ then $\Phi(x+y) = \Phi((\lambda+1)y) = (\lambda+1)^2\Phi(y)$ and $\Phi(x) = \lambda^2\Phi(y)$