ta có \(P=3-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{4}{c}\) theo bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}=\dfrac{4}{6-c}\Rightarrow-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\le-\dfrac{4}{6-c}=\dfrac{4}{c-6}\)
\(\Rightarrow P\le3+\dfrac{4}{c-6}-\dfrac{4}{c}\)\(=3+\dfrac{24}{c^2-6c}\)
\(\Rightarrow P\) lớn nhất khi \(\dfrac{24}{c^2-6c}\) lớn nhất
\(\Leftrightarrow c^2-6c\) nhỏ nhất mà \(c^2-6c=c^2-6c+9-9=\left(c-3\right)^2-9\ge-9\)
\(\Rightarrow c^2-6c\ge-9\) \(\Rightarrow\dfrac{24}{c^2-6c}\le\dfrac{-24}{9}\)
\(\Rightarrow P\le3-\dfrac{24}{9}=\dfrac{1}{3}\)\(\Rightarrow MaxP=\dfrac{1}{3}\) dấu bằng xảy ra tại \(c=3;a=b=\dfrac{3}{2}\)
\(P=1-\dfrac{1}{a}+1-\dfrac{1}{b}+1-\dfrac{4}{c}=3-\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{4}{c}\right)\)
Áp dụng BĐT Cauchy-Shwarz dạng Engel, ta có:
\(P\le3-\dfrac{\left(1+1+2\right)^2}{a+b+c}=3-\dfrac{16}{6}=\dfrac{1}{3}\)
Vậy: \(M\text{ax}_P=\dfrac{1}{3}\Leftrightarrow\)\(\left\{{}\begin{matrix}a=b=\dfrac{3}{2}\\c=3\end{matrix}\right.\)
