How prove this interesting identity $(y_{1})^2\cdot y_{2}\cdot y_{3}=x^2_{1}\cdot x_{2}\cdot x_{3}$ 
let $0<x_{1}<x_{2}<x_{3}$, and there exsit $a$ such
  $$\begin{cases}
y_{1}=x_{1}-\ln{x_{1}}=\dfrac{x^2_{1}}{ax_{1}+\ln{x_{1}}}\\
y_{2}=x_{2}-\ln{x_{2}}=\dfrac{x^2_{2}}{ax_{2}+\ln{x_{2}}}\\
y_{3}=x_{3}-\ln{x_{3}}=\dfrac{x^2_{3}}{ax_{3}+\ln{x_{3}}}
\end{cases}$$
show that:
  $$(y_{1})^2\cdot y_{2}\cdot y_{3}=x^2_{1}\cdot x_{2}\cdot x_{3}$$

My try: since
$$(y_{1})^2\cdot y_{2}\cdot y_{3}=x^2_{1}\cdot x_{2}\cdot x_{3}$$
$$\Longleftrightarrow \dfrac{x^4_{1}x^2_{2}x^2_{3}}{(ax_{1}+\ln{x_{1}})(ax_{2}+\ln{x_{2}})(ax_{3}+\ln{x_{3}})}=x^2_{1}\cdot x_{2}\cdot x_{3}$$
$$\Longleftrightarrow (ax_{1}+\ln{x_{1}})(ax_{2}+\ln{x_{2}})(ax_{3}+\ln{x_{3}})=x^2_{1}\cdot x_{2}\cdot x_{3}$$
since
$$\ln{x_{1}}=x_{1}-y_{1},\ln{x_{2}}=x_{2}-y_{2},\ln{x_{3}}=x_{3}-y_{3}$$
then
$$\Longleftrightarrow [(a+1)x_{1}-y_{1}][(a+1)x_{2}-y_{2}][(a+1)x_{3}-y_{3}]=x^2_{1}\cdot x_{2}\cdot x_{3}$$
then I can't Continue ,This problem is from china's college entrance examination simulation.
and this problem not have  solution
Thank you
 A: Note $x_i>0$, and since $e^x \geq 1+x$ we have $y_i=x_i-\ln x_i \geq 1$. We have 
$$y_i=x_i-\ln x_i=\frac{x_i^2}{ax_i+\ln x_i}=\frac{x_i^2}{ax_i+x_i-y_i}=\frac{x_i^2}{(a+1)x_i-y_i}$$
$$(a+1)x_iy_i-y_i^2=x_i^2$$
Let $a+1=2b$, so $2bx_iy_i-y_i^2=x_i^2$.
$$(x_i-by_i)^2=(b^2-1)y_i^2$$
$$x_i=\left(b \pm \sqrt{b^2-1}\right)y_i$$

Lemma: The equation $\frac{x}{x-\ln x}=c$ has exactly one positive real root if $0<c \leq 1$, and at most two positive real roots if $1<c$.
Proof: Let $f(x)=\frac{x}{x-\ln x}$. We see $f'(x)=\frac{1-\ln x}{(x-\ln x)^2}$, so $f$ is strictly increasing on $(0, e]$, has a maximum at $x=e$, and is strictly decreasing on $[e, \infty)$. Furthermore $f(e)=\frac{e}{e-1}>1$, $\lim_{x \to 0}{f(x)}=0$ and $\lim_{x \to \infty}{f(x)}=1$.
It follows that $f$ restricted to $(0, e]$ is a strictly increasing bijection to $(0, \frac{e}{e-1}]$ and $f$ restricted to $[e, \infty)$ is a strictly decreasing bijection to $(1, \frac{e}{e-1}]$. Now the desired conclusion follows easily.

Case 1: $x_i=(b-\sqrt{b^2-1})y_i$.
Since $b-\sqrt{b^2-1} \leq 1$, we have $x_i \leq y_i=x_i-\ln x_i$ so $x_i \leq 1$.
Note $\frac{x_i}{x_i-\ln x_i}=b-\sqrt{b^2-1}$. By the lemma applied to $0<c=b-\sqrt{b^2-1} \leq 1$, we have at most one possible solution for $x_i$.

Case 2: $x_i=(b+\sqrt{b^2-1})y_i$.
Since $b+\sqrt{b^2-1} \geq 1$, we have $x_i \geq y_i=x_i-\ln x_i$ so $x_i \geq 1$. 
Note $\frac{x_i}{x_i-\ln x_i}=b+\sqrt{b^2-1}$. By the lemma applied to $1<c=b+\sqrt{b^2-1}$, we have at most two possible solutions for $x_i$.

Since $0<x_1<x_2<x_3$, we must have $x_1$ satisfying Case 1 and $x_2, x_3$ satisfying case 2.
Thus 
$$x_1=(b-\sqrt{b^2-1})y_1$$
$$x_2=(b+\sqrt{b^2-1})y_2$$
$$x_3=(b+\sqrt{b^2-1})y_3$$
Thus
$$x_1^2x_2x_3=(b-\sqrt{b^2-1})^2(b+\sqrt{b^2-1})(b+\sqrt{b^2-1})y_1^2y_2y_3=y_1^2y_2y_3$$
