
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Determine the possible values of \( x \) and \( y \):

\[1-\sqrt{x-1}=\sqrt{y-1}\]

From which we get \(\left\{\begin{array}{c}x \geq 1, \\ 1-\sqrt{x-1} \geq 0 .\end{array} \rightarrow x \in[1,2]\right.\). Similarly, \( y \in[1,2] \).

Solve the equation:

\[1-\sqrt{x-1}=\sqrt{y-1}, \rightarrow y-1=1-2 \sqrt{x-1}+x-1 \text {, i.e. } y=1-2 \sqrt{x-1}+x\]

Let \(\sqrt{x-1}=t\), then we have \(\left\{\begin{array}{c}x=t^{2}+1, \\ y=1-2 t+t^{2}+1 .\end{array} \rightarrow x+y=2 t^{2}-2 t+3\right.\).

Since \( x \in[1,2] \), then \( t \in[0,1] \). Thus, we are led to study the quadratic function \( f(t)=2 t^{2}-2 t+3 \) to find its maximum and minimum values on the interval \( t \in[0,1] \). Express the function as

\[f(t)=2\left(t-\frac{1}{2}\right)^{2}+\frac{5}{2}\]

The graph of this function is a parabola opening upwards with the vertex at the point \(\left(\frac{1}{2} ; \frac{5}{2}\right)\). Therefore, the minimum value occurs at \( t=\frac{1}{2} \) (for \( x=y=\frac{5}{4} \)), and the maximum value occurs at \( t=0 \) and \( t=1 \), \( f(0)=f(1)=3 \) (for \( x=2, y=1 \) or \( x=1, y=2 \)).

Present the answer in LaTex format: \boxed{Your answer}
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