chém bài 4
có [TEX]\frac{1}{a+2b+3c}\le \frac{1}{6^2}(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})[/TEX]
tương tự [TEX]\frac{1}{2a+3b+c}\le \frac{1}{6^2}(\frac{2}{a}+\frac{3}{c}+\frac{1}{c})[/TEX]
[TEX]\frac{1}{3a+b+2c}\le \frac{1}{6^2}(\frac{3}{a}+\frac{1}{b}+\frac{2}{c})[/TEX]
cộng lại đc [TEX]LHS \le \frac{1}{6^2}(\frac{6}{a}+\frac{6}{b}+\frac{6}{c})[/TEX]
[TEX]=\frac{1}{6}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})[/TEX]
[TEX]=\frac{1}{6}(\frac{ab+bc+ca}{abc}) =\frac{1}{6} <\frac{3}{6}[/TEX]