value of expression having variables $p,q,r,x,y,z$

Question

If $p,q,r,x,y,z$ are non zero real number such that

$px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}=0$

Then $\displaystyle \frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}$ is

what try

$(px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2)$

$p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2y^2+p^2z^2+q^2x^2+q^2y^2+q^2z^2+r^2x^2+r^2y^2+r^2z^2$

$2pqxy+2qryz+2prxz=p^2y^2+p^2z^2+q^2x^2+q^2z^2+r^2x^2+r^2y^2$

How do i solve it Help me please

Answer 1

Write $v=(p,q,r)$ and $w=(x,y,z)$. Then the given relation states $$v\cdot w+|v||w|=0$$ But $v\cdot w=|v||w|\cos\theta$ where $\theta$ is the angle between them, so $$|v||w|(\cos\theta+1)=0$$ Since none of the scalars are zero, we get $\cos\theta=-1$, so $v=kw$ for some nonzero $k\in\mathbb R$ and $$\frac{py}{qx}+\frac{qz}{ry}+\frac{rx}{pz}=\frac{kxy}{kyx}+\frac{kyz}{kzy}+\frac{kzx}{kxz}=3$$

Answer 2

By C-S $$0=px+qy+rz+\sqrt{(p^2+q^2+r^2)(x^2+y^2+z^2)}\geq px+qy+rz+|px+qy+rz|$$ and since $$px+qy+rz+|px+qy+rz|\geq0,$$ we obtain $$px+qy+rz+|px+qy+rz|=0,$$ which gives $$px+qy+rz\leq0.$$ Also, the equality occurs for $$(x,y,z)||(p,q,r),$$ which says that there is $k<0$, for which $$(p,q,r)=k(x,y,z).$$ Thus, $$\sum_{cyc}\frac{py}{qx}=\sum_{cyc}\frac{kxy}{kyx}=3.$$