\(\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)
\(\Leftrightarrow\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=8\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=8abc\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ac+c^2\right)-8abc=0\)
\(\Leftrightarrow a^2b+abc+a^2c+ac^2+ab^2+b^2c+abc+bc^2-8abc=0\)
\(\Leftrightarrow\left(a^2b-2abc+c^2b\right)+\left(a^2c-2abc+b^2c\right)+\left(ab^2-2abc+ac^2\right)=0\)
\(\Leftrightarrow b\left(a-c\right)^2+c\left(a-b\right)^2+a\left(b-c\right)^2=0\)
Do a;b;c dương nên \(b\left(a-c\right)^2;c\left(a-b\right)^2;a\left(b-c\right)^2\ge0\forall a;b;c\)
\(\Rightarrow b\left(a-c\right)^2+c\left(a-b\right)^2+a\left(b-c\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c\) Thay vào P ta được :
\(P=\frac{a^3+a^3+a^3}{a.a.a}=\frac{3a^3}{a^3}=3\)
