Nhận xét: Với x,y > 0 ta có:
\(4xy\le\left(x+y\right)^2\)
<=> \(\dfrac{1}{x+y}\le\dfrac{x+y}{4xy}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Xảy ra khi x = y
Áp dụng và bài ta có:
\(\dfrac{1}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{2a}+\dfrac{1}{b+c}\right)\le\dfrac{1}{4}\left[\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\right]=\dfrac{1}{8}\left(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{2c}\right)\)
Tương tự: \(\dfrac{1}{a+2b+c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2c}\right)\);
\(\dfrac{1}{a+b+2c}\le\dfrac{1}{8}\left(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{c}\right)\)
Cộng 3 vế bđt có:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{3}{4}\)
