Question

Cho \(a^3+b^3+c^3=3abc\)

\(\left(a;b;c\right)>0\) và khác 0

Tính \(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{a}{c}\right)\)

Answer 1

Áp dụng bđt AM-GM cho 2 số dương:

\(a^3+b^3+c^3\ge3abc\)

Dấu "=" xảy ra khi:

\(a=b=c\)

Khi đó:

\(\left\{{}\begin{matrix}\dfrac{a}{b}=1\\\dfrac{b}{c}=1\\\dfrac{a}{c}=1\end{matrix}\right.\) \(\Leftrightarrow\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Answer 2

Ta có: \(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a+b+c=0\) hoặc \(a=b=c\) (bn tự chứng minh)

+) \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;a+c=-b\)\(\Rightarrow A=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)

\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)

+) \(a=b=c\Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)