Cho a, b, c dương
Chứng minh: $\frac{a+b+3c}{3a+3b+2c}+\frac{b+c+3a}{3b+3c+2a}+\frac{c+a+3b}{3c+3a+2b}> \frac{15}{8}$
Ta có: $\sum (\frac{a+b+3c}{3a+3b+2c}+2)=7.\sum (\frac{a+b+c}{3a+3b+2c})=\frac{7}{8}.((3a+3b+2c)+(3a+2b+3c)+(2a+3b+3c)).(\frac{1}{3a+3b+2c}+\frac{1}{3a+2b+3c}+\frac{1}{2a+3b+3c})\geq \frac{7}{8}.9=\frac{63}{9}\Rightarrow$VẾ TRÁI$\geq \frac{63}{9}-6=\frac{15}{8}$
Dấu bằng xảy ra $\Leftrightarrow a=b=c$(đề thiếu dấu bằng à bạn)