8. Arrange in order of magnitude the values of the expressions $A, B$ and $C$, if $a \geq 1, b \geq 1, a<b, x \neq y$: (20)

$$
\begin{aligned}
A & =\left[\left(a+\frac{z-x y}{x-y}\right) \cdot\left(a-\frac{z-x y}{x-y}\right)+\left(\frac{z-x y}{x-y}\right)^{2}\right]: b^{2} \\
B & =\left(\frac{a+1}{a+2}+\frac{1}{a}\right):\left(\frac{a+1}{a}-\frac{1}{a+2}\right) \cdot(a+b) \\
C & =\frac{a}{a+\frac{1}{a+\frac{1}{a}}} \cdot \frac{a+\frac{2}{a}}{a+\frac{1}{a}}
\end{aligned}
$$
Solution.

$$
\begin{aligned}
A & =\left[\left(a+\frac{z-x y}{x-y}\right) \cdot\left(a-\frac{z-x y}{x-y}\right)+\left(\frac{z-x y}{x-y}\right)^{2}\right]: b^{2} \\
& =\left[a^{2}-\left(\frac{z-x y}{x-y}\right)^{2}+\left(\frac{z-x y}{x-y}\right)^{2}\right]: b^{2}=\frac{a^{2}}{b^{2}}<1 \\
B & =\left(\frac{a+1}{a+2}+\frac{1}{a}\right):\left(\frac{a+1}{a}-\frac{1}{a+2}\right) \cdot(a+b) \\
& =\frac{a(a+1)+a+2}{a(a+2)}: \frac{(a+2)(a+1)-a}{a(a+2)(a+2)(a+1)-a} \cdot(a+b) \\
& =\frac{a^{2}+2 a+2}{a(a+2)} \cdot \frac{a(a+2)}{a^{2}+2 a+2} \cdot(a+b)=a+b \geq 2
\end{aligned}
$$

(6 points)

(6 points)

$$
C=\frac{a}{a+\frac{1}{a+\frac{1}{a}}} \cdot \frac{a+\frac{2}{a}}{a+\frac{1}{a}}=\frac{a}{a+\frac{a}{a^{2}+1}} \cdot \frac{\frac{a^{2}+2}{a}}{\frac{a^{2}+1}{a}}=\frac{a^{2}+1}{a^{2}+2} \cdot \frac{a^{2}+2}{a^{2}+1}=1
$$

(6 points)

The order is $A<C<B$, i.e., the smallest value is of expression $A$, and the largest is of expression $B$.

(2 points)

## MUNICIPAL/SCHOOL COMPETITION IN MATHEMATICS