Example 1 (Question from the 11th "Hope Cup") Given $0<a<b, x=\sqrt{a+b}-\sqrt{b}, y=\sqrt{b}-\sqrt{b-a}$, then the size relationship between $x, y$ is $\qquad$ .
Solution 1: $x=\sqrt{a+b}-\sqrt{b}=\frac{a}{\sqrt{a+b}+\sqrt{b}}, y=\sqrt{b}-\sqrt{b-a}=\frac{a}{\sqrt{b}+\sqrt{b-a}}$.
Since $0<\sqrt{b}+\sqrt{b-a}$,
it follows that $x<\frac{a}{\sqrt{b}+\sqrt{b-a}}=y$,
and since $a>0$ and $b-a>0$, we have $\sqrt{a+b}+\sqrt{b}>\sqrt{b}+\sqrt{b-a}$,
so $\frac{x}{y}<1$,
thus $x<y$.

Solution 3: The original problem is equivalent to comparing $\sqrt{a+b}+\sqrt{b-a}$ with $2\sqrt{b}$. Since $\sqrt{a+b}+\sqrt{b-a}=\sqrt{(a+b)+2\sqrt{b^2-a^2}+(b-a)}=\sqrt{2b+2\sqrt{b^2-a^2}}<\sqrt{2b+2\sqrt{b^2}}=2\sqrt{b}$, it follows that $x<y$.

Solution 4: As shown in Figure 1-1, take three different points $A(b-a, \sqrt{b-a})$, $B(b, \sqrt{b})$, and $C(a+b, \sqrt{a+b})$ on the graph of the function $y=\sqrt{x}$. From the graph, it is clear that $k_{BC}<k_{AB}$,
i.e., $\frac{\sqrt{a+b}-\sqrt{b}}{(a+b)-b}<\frac{\sqrt{b}-\sqrt{b-a}}{b-(b-a)}$,
which means $\sqrt{a+b}-\sqrt{b}<\sqrt{b}-\sqrt{b-a}$,
thus $x<y$.