Lời giải:
Ta có: Với mọi \(a,b,c\in\mathbb{R}\) thì
\(\left\{\begin{matrix} (2ab-1)^2\geq 0\\ (3bc-2)^2\geq 0\\ (4ac-3)^2\geq 0\end{matrix}\right.\Rightarrow (2ab-1)^2+(3bc-2)^2+(4ac-3)^2\geq 0\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} 2ab-1= 0\\ 3bc-2= 0\\ 4ac-3= 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} ab=\frac{1}{2}\\ bc=\frac{2}{3}\\ ac=\frac{3}{4}\end{matrix}\right.\)
\(\Rightarrow (abc)^2=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}=\frac{1}{4}\)\(\Leftrightarrow abc=\pm \frac{1}{2}\)
Nếu \(abc=\frac{1}{2}\)
\(\Rightarrow \left\{\begin{matrix} c=\frac{abc}{ab}=\frac{1}{2}:\frac{1}{2}=1\\ a=\frac{abc}{bc}=\frac{1}{2}:\frac{2}{3}=\frac{3}{4}\\ b=\frac{abc}{ac}=\frac{1}{2}:\frac{3}{4}=\frac{2}{3}\end{matrix}\right.\)
Nếu \(abc=\frac{-1}{2}\). Tương tự như trên ta có:
\(\left\{\begin{matrix} c=-1\\ a=\frac{-3}{4}\\ b=\frac{-2}{3}\end{matrix}\right.\)
Vậy......
\(\left(2ab-1\right)^2+\left(3bc-2\right)^2+\left(4ac-3\right)^2\ge0\forall x;b;c\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}2ab=1\\3bc=2\\4ac=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ab=\dfrac{1}{2}\\bc=\dfrac{2}{3}\\ac=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left(abc\right)^2=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}=\dfrac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}abc=\dfrac{1}{2}\\abc=-\dfrac{1}{2}\end{matrix}\right.\)
Đến đây tự tìm được \(a;b;c\)
