Áp dụng BĐT sau: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
\(\Rightarrow\frac{1}{2a+b+c}\le\frac{1}{4}\left(\frac{1}{2a}+\frac{1}{b+c}\right)\). Lại có \(\frac{1}{b+c}\le\frac{1}{4b}+\frac{1}{4c}\)
\(\Rightarrow\frac{1}{2a+b+c}\le\frac{1}{4}\left(\frac{1}{2a}+\frac{1}{4b}+\frac{1}{4c}\right)\)
Tương tự: \(\frac{1}{a+2b+c}\le\frac{1}{4}\left(\frac{1}{4a}+\frac{1}{2b}+\frac{1}{4c}\right);\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{4a}+\frac{1}{4b}+\frac{1}{2c}\right)\)
Cộng 3 BĐT trên theo vế, ta được:
\(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Thay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4\)\(\Rightarrow\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le1\)(đpcm).
Dấu "=" xảy ra <=> \(a=b=c=\frac{3}{4}.\)
