\(M=4x^2-3x+\dfrac{1}{4x}+2011\)
\(M=4x^2-4x+1+x+\dfrac{1}{4x}+2011\)
\(M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\)
Vì \(\left(2x-1\right)^2\ge0\) và \(x>0\)
\(\Rightarrow\dfrac{1}{4x}>0\)
Lợi dụng BĐT Cauchy cho 2 số nguyên dương ta có:
\(x+\dfrac{1}{4x}\ge2\sqrt{x\dfrac{1}{4x}}=2.\dfrac{1}{2}=1\)
\(\Rightarrow M=\left(2x-1\right)^2+\left(x+\dfrac{1}{4x}\right)+2010\ge0+1+2010=2011\)
\(\Rightarrow M\ge2011\)
Dấu " = " xảy ra khi:
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\x=\dfrac{1}{4x}\\x>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x^2=\dfrac{1}{4}\\x>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\\x>0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(M_{min}=2011\) đạt được khi \(x=\dfrac{1}{2}\)
