Find $\frac{a+b+c}{x+y+z}$ given $a^2+b^2+c^2$, $x^2+y^2+z^2$ and $ax+by+cz$. We are given $a^2+b^2+c^2=m$, $x^2+y^2+z^2=n$ and $ax+by+cz=p$ where $m,n$ and $p$ are known constants. Also, $a,b,c,x,y,z$ are non-negative numbers.
The question asks to find the value of $\dfrac{a+b+c}{x+y+z}$.
I have thought a lot about this problem, but I can't solve it.
I can write $(a+b+c)^2$ as $a^2+b^2+c^2+2(ab+bc+ca)=m+2(ab+bc+ca)$, but then I don't know how to determine $ab+bc+ca$. I'm also not certain how to use the equation $ax+by+cz=p$ to help me in finding the given expression's value. 
Any help would be much appreciated.
 A: I don't think this can be solved. Think of $A=(a, b, c)$ and $X=(x, y, z)$ as vectors. Then we know $A\cdot A$, $X\cdot X$ and $A\cdot X$. So all we know is the lengths of $A$ and $X$, and the angle between them. Now let $C=(1,1,1)$. What we want is
$$\frac{A\cdot C}{X\cdot C}$$
But if we hold $A$ and $C$ fixed we can still vary $X$ whilst keeping $X\cdot X$ and $A\cdot X$ the same: Just rotate $X$ around the axis given by $A$. This changes $X\cdot C$ without changing $A\cdot C$ and so it has to change the value of the expression we want to calculate. So this expression is not determined purely by the quantities given.
A: Let $m=5/4,n=2,p=1$. We can produce a one parameter family of solutions to the three equations so that $(a+b+c)/(x+y+z)$ has infinitely many values for that family.
Let $a=\cos t,\ b=\sin t, c=1/2$ (so $a^2+b^2+c^2=5/4$ is satisfied.)
Let $x=\cos u,\ y=\sin u, z=1$ (so $x^2+y^2+z^2=2$ is satisfied.)
Using the difference formula for cosine, the third equation becomes
$$\cos(t-u)+1/2=1,$$
which holds as long as say $t-u=\pi/3$. So put $t=u+\pi/3$ [with also $0<u<\pi/6$ to keep $t,u$ in the first quadrant so that $a,b,x,y$ are positive] and consider the quantity $(a+b+c)/(x+y+z)$, which becomes
$$\frac{\cos(u+\pi/3)+\sin(u+\pi/3)+1/2}{\cos u + \sin u + 1},$$
which is a smooth nonconstant function of $u$ and so takes on infinitely many values.
