ta có a > 0 → b + c < 1
→ 4bc < (b + c)² < 1
→ bc < 1\4
tương tự với ab, ac là => dpcm
ta có a > 0 → b + c < 1
→ 4bc < (b + c)² < 1
→ bc < 1\4
tương tự với ab, ac là => dpcm
\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}=\frac{ab}{\left(a+c\right)+\left(b+c\right)}+\frac{bc}{\left(a+b\right)+\left(a+c\right)}+\frac{ac}{\left(a+b\right)+\left(b+c\right)}\)(vì a+b+c=1)
Theo Cauchy : \(a+b\ge2\sqrt{ab}\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow\frac{a+b}{4ab}\ge\frac{1}{a+b}\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Áp dụng vào BĐT ta đc:
\(\frac{ab}{\left(a+c\right)+\left(c+b\right)}+\frac{bc}{\left(a+b\right)+\left(a+c\right)}+\frac{ac}{\left(a+b\right)+\left(b+c\right)}\)
\(\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)+\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{ac}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)
\(\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)
\(\le\frac{1}{4}\left(\frac{ab+bc}{a+c}+\frac{ab+ac}{b+c}+\frac{bc+ac}{a+b}\right)=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)
Dấu''='' xảy ra khi a=b=c=1/3
