đặt 6a=x;2b=y;3c=z=>x+y+z=11
áp dụng bất đẳng thức Schwarts ta có:\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+y+z+3}=\frac{9}{14}\)
\(\Leftrightarrow\frac{28}{x+1}+\frac{28}{y+1}+\frac{28}{z+1}\ge\frac{28.9}{14}=18\)
\(\Leftrightarrow\frac{28}{x+1}-1+\frac{28}{y+1}-1+\frac{28}{z+1}-1\ge18-1-1-1=15\)
\(\Leftrightarrow\frac{27-x}{x+1}+\frac{27-y}{y+1}+\frac{27-z}{z+1}\ge15\)
\(\Leftrightarrow\frac{11-x+16}{x+1}+\frac{11-y+16}{y+1}+\frac{11-z+16}{z+1}\ge15\)
\(\Leftrightarrow\frac{y+z+16}{x+1}+\frac{z+x+16}{y+1}+\frac{x+y+16}{z+1}\ge15\)
\(\Leftrightarrow\frac{2b+3c+16}{6a+1}+\frac{6a+3c+16}{2b+1}+\frac{6a+2b+16}{3c+1}\ge15\)
=>đpcm
dấu "=" xảy ra khi \(a=\frac{11}{18};b=\frac{11}{6};c=\frac{11}{9}\)
