Áp dụng bđt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Ta có
\(\frac{1}{2a+b+c}\le\frac{1}{4}\left(\frac{1}{2a}+\frac{1}{b+c}\right)=\frac{1}{8a}+\frac{1}{16}\left(\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{a+2b+c}\le\frac{1}{8b}+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{c}\right)\)
\(\frac{1}{a+b+2c}\le\frac{1}{8c}+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\sum\frac{1}{2a+b+c}=\sum\frac{1}{a+a+b+c}\le\frac{1}{16}\sum\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{3}{4}\)