Find all non-zero real numbers $x,y,z$ which satisfy the system of equations:
\begin{align}
(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)&=xyz,\\(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)(z^4+z^2x^2+x^4)&=x^3y^3z^3
\end{align}
It's an indian olympiad question. Can you guys help me out in solving it please ?
edit:
I have tried to write the first equation as:
$(\frac{x^3-y^3}{x-y})(\frac{y^3-z^3}{y-z})(\frac{z^3-x^3}{z-x})$ = xyz
And second equation as:
$(\frac{x^6-y^6}{x^2-y^2})(\frac{y^6-z^6}{y^2-z^2})(\frac{z^6-x^6}{z^2-x^2}) = x^3y^3z^3$
Then divided the two to get:
$(\frac{x^3+y^3}{x+y})(\frac{y^3+z^3}{y+z})(\frac{z^3+x^3}{z+x}) = x^2y^2z^2$
After that I have no sign what to do??
Best Answer
You obtained $$\left(\frac{x^3+y^3}{x+y}\right)\!\!\left(\frac{y^3+z^3}{y+z}\right)\!\!\left(\frac{z^3+x^3}{z+x}\right) = x^2y^2z^2$$ which is equivalent to $$\prod_\mathrm{cyc}(x^2-xy+y^2)=x^2y^2z^2.$$ Dividing throughout by $xy\cdot yz\cdot zx$, we get $$\prod_{\mathrm{cyc}}\left(\frac{x}{y}-1+\frac{y}{x}\right)=1.$$ Now, we use the hint given by achille hui in the comments. For every nonzero $u\in\mathbb R$, we have $|u+u^{-1}-1|\geq 1$ with equality exactly when $u=1$. To see this, note that if $u>0$, then $u+u^{-1}\geq2$ by AM-GM inequality, whereas if $u<0$ then by the same logic, $u+u^{-1}\leq -2$. Now, it follows from the multiplicity of $|\cdot|$ that $$\prod_{\mathrm{cyc}}\left|\frac{x}{y}-1+\frac{y}{x}\right|=1.$$ If any one of the terms in the cyclic product is $>1$, then that forces one of the other terms to be $<1$, which we know to be impossible. This implies that all the terms are exactly one, so that $x/y=1$ and $x=y$. Similarly $y=z$, so $x=y=z$. The rest is easy.