Bài lớp 8 thật hả? :(
\(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le1\)
\(\Leftrightarrow\frac{a}{4-a}+\frac{b}{4-b}+\frac{c}{4-c}\le1\)
\(\Leftrightarrow a\left(4-b\right)\left(4-c\right)+b\left(4-a\right)\left(4-c\right)+c\left(4-a\right)\left(4-b\right)\le\left(4-a\right)\left(4-b\right)\left(4-c\right)\)
\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le4\) (1)
Ta cần chứng minh (1)
Không mất tính tổng quát, giả sử \(a\le c\le b\)
\(\Rightarrow a\left(a-c\right)\left(b-c\right)\le0\)
\(\Leftrightarrow a^2b+ac^2\le a^2c+abc\)
\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le a^2c+abc+b^2c+abc\)
\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le c\left(a+b\right)^2\)
\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le\frac{1}{2}.2c\left(a+b\right)\left(a+b\right)\le\frac{1}{2}.\frac{\left(2c+a+b+a+b\right)^3}{27}\)
\(\Leftrightarrow a^2b+ac^2+b^2c+abc\le\frac{1}{2}.\frac{8.3^3}{27}=4\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)