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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 \(a=\frac{11}{18};b=\frac{11}{6};c=\frac{11}{9}\)
Ta có:
A = \(\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{3b+2a}=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{3bc+2ac}\)
A \(\ge\frac{\left(a+b+c\right)^2}{2ab+3ac+2bc+3ab+3bc+2ac}\)(bđt svacxo \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\))
A \(\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ac\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{5\left(a+b+c\right)^2}{3}}\) (bđt \(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\)(*)
CM bđt * <=> \(3xy+3yz+3xz\le x^2+y^2+z^2+2xz+2xy+2yz\)
<=> \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\ge0\) (luôn đúng)
<=> A \(\ge\frac{3}{5}\) --> ĐPCM