Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{}\text{}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)
Ta thấy: \(\text{}\text{}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)
\(\text{}\text{}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)
Do đó: P \(\ge2+4+8=14\)
Vậy: P(min)=14 khi: \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)
làm lại:
Ta có: P= \(2a+3b+\dfrac{4}{a}+\dfrac{9}{b}\) = \(\text{}\text{}(\dfrac{4}{a}+a)+\left(\dfrac{9}{b}+b\right)+\left(a+2b\right)\)
Ta thấy: \(\text{}\text{}(\dfrac{4}{a}+a)\ge2\sqrt{\dfrac{4}{a}\cdot a}=4\)
\(\text{}\text{}\left(\dfrac{9}{b}+b\right)\ge2\sqrt{\dfrac{9}{b}\cdot b}=6\)
Do đó: P \(\ge4+6+8=18\)
Vậy: P(min)=18 khi: \(\left\{{}\begin{matrix}\dfrac{4}{a}=a\\\dfrac{9}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\cdot\)
