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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.

Alraxite
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    Are you given the specific values of $m, n, p$? For example, if $mn = p^2$, that tells us a lot more, than just saying that they are (random) known constant. – Calvin Lin Oct 08 '13 at 00:53
  • @CalvinLin Isn't that the only case when the solution is unique? – N. S. Oct 08 '13 at 03:31
  • @CalvinLin Thank you for pointing that out. I saw this problem somewhere where instead of $m,n,p$, some specific numbers were indeed given. I didn't remember the numbers, so I changed them to arbitrary constants in my post. I didn't realize that this would considerably change the solution. I will try to find the original problem with the actual numbers and if I'm not able to solve it, I will post it as a new question (since that would require a completely different solution). – Alraxite Oct 08 '13 at 04:25
  • @N.S. Indeed. My point is that the question isn't asking for general values of $m,n,p$, but for a specific set of values, which results in the equality condition of Cauchy-Schwarz. – Calvin Lin Oct 08 '13 at 04:46

2 Answers2

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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.

Oscar Cunningham
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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.

coffeemath
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