Lời giải:
\(B=\frac{1}{a}+\frac{1}{b}-\left(\frac{a}{b}+\frac{b}{a}-2\right)=\frac{a+b}{ab}-\frac{a^2+b^2}{ab}+2\)
\(=\frac{a+b-1}{ab}+2=\frac{a+b-1}{\frac{(a+b)^2-(a^2+b^2)}{2}}+2=\frac{2(a+b-1)}{(a+b)^2-1}+2\)
\(=\frac{2}{a+b+1}+2\)
Áp dụng BĐT AM-GM:
\(2=2(a^2+b^2)\geq (a+b)^2\Rightarrow a+b\leq \sqrt{2}\)
\(\Rightarrow B\geq \frac{2}{\sqrt{2}+1}+2=2\sqrt{2}\)
Vậy $B_{\min}=2\sqrt{2}$ tại $a=b=\frac{1}{\sqrt{2}}$