a/ \(B=\dfrac{x^2-x-1}{x^2+x+1}\Leftrightarrow Bx^2+Bx+B=x^2-x-1\)
\(\Leftrightarrow\left(B-1\right)x^2+\left(B+1\right)x+B+1=0\)
\(\Delta=\left(B+1\right)^2-4\left(B-1\right)\left(B+1\right)\ge0\)
\(\Leftrightarrow\left(B+1\right)\left(B+1-4B+4\right)\ge0\)
\(\Leftrightarrow\left(B+1\right)\left(5-3B\right)\ge0\)
\(\Rightarrow-1\le B\le\dfrac{5}{3}\) \(\Rightarrow\left[{}\begin{matrix}B_{max}=\dfrac{5}{3}\\B_{min}=-1\end{matrix}\right.\)
b/ Áp dụng BĐT Cô-si:
\(\left\{{}\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\end{matrix}\right.\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
