\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a^3+b^3+3a^2b+3b^2a\right)+c^3-3a^2b-3b^2a-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\left(1\right)\)
C/m : \(a^2+b^2+c^2-ab-bc-ac\ge0\)
Giả sử điều phải c/m là đúng , ta có :
\(a^2+b^2+c^2-ab-bc-ac\ge0\)
\(\Rightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)\ge0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ( điều này luôn đúng )
\(\Rightarrow\) điều giả sử là đúng
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac\ge0\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow a+b+c=0\)
\(\Rightarrow a+b=-c;b+c=-a;a+c=-b\)
Lại có : \(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)\)
\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)
\(=\dfrac{-abc}{abc}=-1\)
Vậy \(A=-1\)