Question

Prove that the equation x/y + y/z + z/x = 1 has no solutions in positive integers x,y,z.​

Answer 1

Required answer.

By A.M-G.M inequality, $\dfrac{x}{y} +\dfrac{y}{z} +\dfrac{z}{x} \geq 3\sqrt[3]{\dfrac{x}{y} \times \dfrac{y}{z} \times \dfrac{z}{x} } =3$ for $x,y,z> 0$. Hence the value of $\dfrac{x}{y} +\dfrac{y}{z} +\dfrac{z}{x}$ must be greater than 1, hence no solution to the given equation.

Proof.

Prove that $a^3+b^3+c^3\geq 3abc$ for $a,b,c> 0$.

$a^3+b^3+c^3-3abc$

$= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\dfrac{1}{2} (a+b+c)\{(a-b)^2+(b-c)^2+(c-a)^2\}\geq 0$since $a+b+c>0$. And hence $a^3+b^3+c^3\geq 3abc$. Equality holds where $a=b=c$.

Answer 2

Hence the value of $\dfrac{x}{y} +\dfrac{y}{z} +\dfrac{z}{x}$ must be greater than 1, hence no solution to the given equation.

Proof.

Prove that $a^3+b^3+c^3\geq 3abc$ for $a,b,c> 0$.

$a^3+b^3+c^3-3abc$

$= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\dfrac{1}{2} (a+b+c)\{(a-b)^2+(b-c)^2+(c-a)^2\}\geq 0$since $a+b+c>0$. And hence $a^3+b^3+c^3\geq 3abc$. Equality holds where $a=b=c$.