\(题目：已知a，b>0,则\frac{a^2+b^2}{\sqrt{ab-a-b+1}}的最小值为()\)
\(解 ：\)
\(原式=\frac{a^2+b^2}{\sqrt{ab-a-b+1}}\)
\(\quad =\frac{a^2+b^2}{\sqrt{a(b-1)-(b-1)}}\)
\(\quad =\frac{a^2+b^2}{\sqrt{(b-1)(a-1)}}\)
设m=b-1>0，n=a-1>0,则a=n+1,b=m+1,
\(原式=\frac{(n+1)^2+(m+1)^2}{\sqrt{mn}}\)
\(\quad\ge \frac{2(n+1)(m+1)}{\sqrt{mn}}\)
\(\quad=2(\frac{nm+n+m+1}{\sqrt{mn}})\)
\(\quad=2(\sqrt{mn}+\frac{\sqrt{n}}{\sqrt{m}}+\frac{\sqrt{m}}{\sqrt{n}}+\frac{1}{\sqrt{mn}})\)
\(\quad=2(\sqrt{mn}+\frac{1}{\sqrt{mn}}+\frac{\sqrt{n}}{\sqrt{m}}+\frac{\sqrt{m}}{\sqrt{n}})\)
\(\quad \ge 2(2+2)\)
\(\quad \ge 8\)