Bài 6 :
Áp dụng Bđt Cauchy cho 2 số \(x;y>0\)
\(x+y\ge2\sqrt{xy}\Rightarrow4xy\le\left(x+y\right)^2\)
\(\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{x+y}{4xy}\)
\(\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Áp dụng vào đề bài ta có :
\(\dfrac{1}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{2a}+\dfrac{1}{b+c}\right)\le\dfrac{1}{4}\left[\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\right]=\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c}\right)\left(1\right)\)
Tương tự : \(\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\left(2\right)\)
\(\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\left(3\right)\)
\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow M\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Giả sử \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=4\Rightarrow M\le1\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{4}\) thay vào điều kiện đề bài ta được
\(VT=12+13\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=12+13.3.\dfrac{4}{3}=64\)
\(VP=12\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=12.3.\dfrac{16}{9}=64=VP\left(thỏa.mãn\right)\)
Vậy \(GTLN\left(M\right)=1\left(a=b=c=\dfrac{3}{4}\right)\)
