\(B=\frac{2001}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right].\left[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right]\)
\(B=\frac{2001}{2}\left[\frac{a+b}{a+b}+\frac{a+b}{b+c}+\frac{a+b}{c+a}+\frac{b+c}{a+b}+\frac{b+c}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+\frac{c+a}{b+c}+\frac{c+a}{c+a}\right]\)
\(B=\frac{2001}{2}\left[1+\left(\frac{a+b}{b+c}+\frac{b+c}{a+b}\right)+\left(\frac{a+b}{c+a}+\frac{c+a}{a+b}\right)+1+\left(\frac{b+c}{c+a}+\frac{c+a}{b+c}\right)+1\right]\)
Dễ dàng chứng minh được \(\frac{x}{y}+\frac{y}{x}\ge2\). Suy ra:
\(\left(\frac{a+b}{b+c}+\frac{b+c}{a+b}\right)\ge2\); \(\left(\frac{a+b}{c+a}+\frac{c+a}{a+b}\right)\ge2\); \(\left(\frac{b+c}{c+a}+\frac{c+a}{b+c}\right)\ge2\)
=> \(B\ge\frac{2001}{2}.\left(3+2+2+2\right)=\frac{18009}{2}\)
Dấu = khi và chỉ khi \(\frac{a+b}{b+c}=\frac{b+c}{a+b}=\frac{c+a}{b+c}\Rightarrow a=b=c\)
vậy Min B = 18009/2
