For positive integers $x, y$, let $m_{a}, m_{b}$ be the arithmetic mean and geometric mean of $x, y$ respectively.
(1) If $m_{a}+m_{b}=y-x$, find $\frac{x}{y}$;
(2) Prove: there exists a unique pair of positive integers $(x, y) (x<y)$ such that $m_{a}+m_{b}=40$.
(1) Solution: From the given, we have $\frac{x+y}{2}+\sqrt{x y}=y-x$, which leads to $\sqrt{x y}=\frac{y-3 x}{2}$. 
From this, we get $9 x^{2}-10 x y+y^{2}=0$, which is
$(9 x-y)(x-y)=0$.
Solving this, we get $\frac{x}{y}=\frac{1}{9}$ or 1 (discard the latter).
(2) Proof: From the given, we have $\frac{x+y}{2}+\sqrt{x y}=40$, which leads to
$(\sqrt{x}+\sqrt{y})^{2}=80$.
Thus, $\sqrt{x}+\sqrt{y}=4 \sqrt{5}$, which means
$\sqrt{5 x}+\sqrt{5 y}=20$.
Therefore, $\sqrt{5 y}=\frac{400+5 y-5 x}{40}$
Since $\frac{400+5 y-5 x}{40} \in \mathbf{Q}$, we know $\sqrt{5 y}$ is an integer. So,
$y=5 m^{2}, x=5 n^{2}(m, n \in \mathbf{N})$.
From this, we deduce $m+n=4$, and when $n<m$, $n=1, m=$
3. Therefore, $x=5, y=45$.
Hence, there exists a unique pair of positive integers $(x, y)$ that satisfies $m_{a}+m_{b}=40$.
(Xin Hui, Tianjin Normal University, 300074)
High 189 Given $\alpha \beta \in\left(0, \frac{\pi}{2}\right), x \in \mathbf{R}, a, b \in \mathbf{R}_{+}$,

and $\square$
$$
\begin{array}{l}
\frac{\sin ^{2} \alpha \cdot \sin ^{2} \beta}{a \sin ^{2} \alpha+b \sin ^{2} \beta}+\frac{\cos ^{2} \alpha \cdot \cos ^{2} \beta}{a \cos ^{2} \alpha+b \cos ^{2} \beta} \\
=\frac{\sin ^{4} x}{a}+\frac{\cos ^{4} x}{b} .
\end{array}
$$

Prove: (1) $\alpha=\beta$;
(2) $\frac{\sin ^{2 n} x}{a^{n-1}}+\frac{\cos ^{2 n} x}{b^{n-1}}=\frac{1}{(a+b)^{n-1}}\left(n \in \mathbf{N}_{+}\right)$.

Proof: (1) Note that
$$
\frac{a \sin ^{2} \alpha+b \sin ^{2} \beta}{\sin ^{2} \alpha \cdot \sin ^{2} \beta}=\frac{a}{\sin ^{2} \beta}+\frac{b}{\sin ^{2} \alpha}
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\text { By the Cauchy-Schwarz inequality, } \\
\left(a \sin ^{2} \beta+b \sin ^{2} \alpha\right)\left(\frac{a}{\sin ^{2} \beta}+\frac{b}{\sin ^{2} \alpha}\right) \geq(a+b)^{2} . \\
\text { Hence } \frac{a \sin ^{2} \beta+b \sin ^{2} \alpha}{(a+b)^{2}} \geq \frac{1}{\frac{a}{\sin ^{2} \beta}+\frac{b}{\sin ^{2} \alpha}} \\
=\frac{\sin ^{2} \alpha \cdot \sin ^{2} \beta}{a \sin ^{2} \alpha+b \sin ^{2} \beta} .
\end{array}
$$

Similarly, $\frac{a \cos ^{2} \beta+b \cos ^{2} \alpha}{(a+b)^{2}} \geq \frac{\cos ^{2} \alpha \cdot \cos ^{2} \beta}{a \cos ^{2} \alpha+b \cos ^{2} \beta}$.
(1) + (2) gives
$\frac{\sin ^{2} \alpha \cdot \sin ^{2} \beta}{a \sin ^{2} \alpha+b \sin ^{2} \beta}+\frac{\cos ^{2} \alpha \cdot \cos ^{2} \beta}{a \cos ^{2} \alpha+b \cos ^{2} \beta}$
$\leq \frac{a\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+b\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)}{(a+b)^{2}}$
$$
=\frac{1}{a+b} \text {. }
$$

Also, by the Cauchy-Schwarz inequality, we have
$$
(a+b)\left(\frac{\sin ^{4} x}{a}+\frac{\cos ^{4} x}{b}\right) \geq\left(\sin ^{2} x+\cos ^{2} x\right)^{2}=1 \text {. }
$$

Thus, $\frac{\sin ^{4} x}{a}+\frac{\cos ^{4} x}{b} \geq \frac{1}{a+b}$.
From the given and combining (3) and (4),
$\frac{\sin ^{2} \alpha \cdot \sin ^{2} \beta}{a \sin ^{2} \alpha+b \sin ^{2} \beta}+\frac{\cos ^{2} \alpha \cdot \cos ^{2} \beta}{a \cos ^{2} \alpha+b \cos ^{2} \beta}$
$$
=\frac{1}{a+b}=\frac{\sin ^{4} x}{a}+\frac{\cos ^{4} x}{b} \text {. }
$$

Therefore, by the equality condition of the Cauchy-Schwarz inequality, we get
$$
\frac{\frac{a}{\sin ^{2} \beta}}{a \sin ^{2} \beta}=\frac{\frac{b}{\sin ^{2} \alpha}}{b \sin ^{2} \alpha} \Rightarrow \sin ^{2} \alpha=\sin ^{2} \beta \text {. }
$$

Since $\alpha \in\left(0, \frac{\pi}{2}\right)$, we have $\alpha=\beta$.
Similarly, $\frac{\frac{\sin ^{4} x}{a}}{a}=\frac{\frac{\cos ^{4} x}{b}}{b}=\frac{\sin ^{2} x}{a}=\frac{\cos ^{2} x}{b}$.
Thus, $\sin ^{2} x=\frac{a}{a+b}, \cos ^{2} x=\frac{b}{a+b}$.
(2) Substituting $\sin ^{2} x=\frac{a}{a+b}, \cos ^{2} x=\frac{b}{a+b}$ into the left side of the equation to be proven, we get
Thus, the equation to be proven holds.
(Que Zhengping, Anhui Normal University Affiliated High School, 241001)