Question

Cho a,b,c>0. Tìm min:

\(A=\dfrac{b+c}{-a+b+c}+\dfrac{c+a}{a-b+c}+\dfrac{a+b}{a+b-c}\)

Answer 1

\(chuẩn\) \(hóa\) \(a+b+c=3\) \(\left(do:f\left(ta;tb;tc\right)=f\left(a;b;c\right)\right)\left(0< a;b;c< 3\right)\)

\(\Rightarrow\dfrac{b+c}{-a+b+c}=\dfrac{3-a}{-a+3-a}=\dfrac{3-a}{3-2a}\ge3a-1\)\(\left(1\right)\)

\(thật\) \(vậy\) \(\left(1\right)\Leftrightarrow3-a\ge\left(3a-1\right)\left(3-2a\right)\Leftrightarrow3-a-\left(3-2a\right)\left(3a-1\right)\ge0\Leftrightarrow6\left(a-1\right)^2\ge0\left(đúng\right)\)

\(\Rightarrow A\ge3a-1+3b-1+3c-1=3\left(a+b+c\right)-3=3.3-3=6\)\(\)