Ta có: Q = \(3x+\dfrac{1}{2x}=\dfrac{x}{2}+\dfrac{1}{2x}+\dfrac{5x}{2}\)
Áp dụng bđt cosi cho hai số dương x/2, 1/2x và bđt x \(\ge\)1
Ta có: Q \(\ge2\sqrt{\dfrac{x}{2}\cdot\dfrac{1}{2x}}+\dfrac{5}{2}\cdot1=2\cdot\dfrac{1}{2}+\dfrac{5}{2}=\dfrac{7}{2}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{1}{2x}\\x=1\end{matrix}\right.\) <=> x = 1
Vậy MinQ = 7/2 <=> x = 1