-1

Because it is positive integers so I can multiply both side by $(xy+yz+zx)$ I have tried to use completing square, like below $(3x-3y)^2+(3y-3z)^2+(3z-3x)^2=12x^2+12y^2+12z^2$ But in above form, it didn't help much. Then I tried using AM-GM inequality but get nothing.

Lim Zhao Sen
  • 29
  • Please include the question in the body of the post. – Another User May 01 '24 at 10:41
  • 1
    Solution on AoPS: https://artofproblemsolving.com/community/c6h1958433p13536501 – Martin R May 01 '24 at 11:17
  • If there is a solution in integers, not all zero, to $x^2 + y^2 + z^2 - 3yz - 3zx - 3xy = 0$ then dividing by $\gcd(x,y,z) $ gives a solution with variables not all zero, and $\gcd(x,y,z) = 1,$ variables not all zero. Next: if $x^2 + y^2 + z^2 - 3yz - 3zx - 3xy \equiv 0 \pmod {25},$ then all three $x,y,z \equiv 0 \pmod 5.$ So, this contradicts the assumption of an integer solution with $xyz \neq 0$ This is one way of writing Fermat's Infinite Descent that I like. The quadratic form is thus "anisotropic" over the $5-adic $ numbers. See Cassels, Rational Quadratic Forms – Will Jagy May 01 '24 at 18:07
  • Actually, it follows that the prime $2$ also works; try $\pmod 4$ that may be enough. – Will Jagy May 01 '24 at 18:11

1 Answers1

-1

prove that for any positive integers $(x)$ $(y)$ and $(z)$, the expression $(\frac{x^2 + y^2 + z^2}{xy + yz + zx}$) cannot be equal to 3.

Allow me to provide some preliminaries for this proof:

Lemma 1:(Algebraic Identity) For any real numbers $(a)$, $(b)$, $(c)$, the following identity holds: $a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2)$

Lemma 2: (Cauchy-Schwarz Inequality) For all real numbers $(a)$, $(b)$, $(c)$, $$ (a^2 + b^2 + c^2)(1^2 + 1^2 + 1^2) \geq (a + b + c)^2. $$

{Proof by Contradiction:} Assume, otherwise, that the given expression is equal to 3. $$ \frac{x^2 + y^2 + z^2}{xy + yz + zx} = 3. $$ Multiplying both sides by $(xy + yz + zx)$ $\rightarrow$: $$ x^2 + y^2 + z^2 = 3(xy + yz + zx). $$

Applying Lemma 1, we rearrange the terms: $$ \frac{1}{2}((x-y)^2 + (y-z)^2 + (z-x)^2) = x^2 + y^2 + z^2 - xy - yz - zx. $$ $\rightarrow$: $$ \frac{1}{2}((x-y)^2 + (y-z)^2 + (z-x)^2) = 2(xy + yz + zx). $$ Since each squared term is non-negative, if $(\frac{1}{2}((x-y)^2 + (y-z)^2 + (z-x)^2) = 0)$, it implies $(x = y = z)$. Substituting $(x = y = z)$ into the original assumption: $$ \frac{3x^2}{3x^2} = 1, $$ which contradicts the original assumption.

For the case where $((x-y)^2 + (y-z)^2 + (z-x)^2 > 0)$, the left-hand side remains strictly positive and cannot equate to $(2xy + 2yz + 2zx)$

Thus, by contradiction, $(\frac{x^2 + y^2 + z^2}{xy + yz + zx} \neq 3)$ for any positive integers $(x)$, $(y)$, and $(z).

Sean Wilson
  • 1
  • Why the case where $(x-y)^2+(y-z)^2+(z-x)^2>0$ cannot equate to $2xy+2yz+2zx$ ? I think you should prove that claim. – Lim Zhao Sen May 01 '24 at 13:06
  • Assume the case of $(x-y)^2+(y-z)^2+(z-x)^2 \gt 0$ $\forall$ x,y,z at least 2 must be strictly distinct to hold true. $(x-y)^2 = x^2 - 2xy + y^2$ $(y-z)^2 = y^2 - 2yz + z^2$ $(z-x)^2 = z^2 - 2zx + x^2$

    Consider the term $(x-y)^2$: In order for it to contribute $2xy$ to the sum, we'd need: $x^2 - 2xy + y^2 = 2xy$ Simplifying this gives: $x^2 - 4xy + y^2 = 0$ This would imply $x = 2y$, which contradicts the assumption that x, y are distinct. Similar reasoning can be applied for the rest of the squared difference terms accordingly. If we didn’t have >0 then aye plenty of intersects.

     – Sean Wilson May 02 '24 at 16:06