Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{ab}{c+1}=\frac{ab}{\left(c+a\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right);\frac{ac}{b+1}\le\frac{1}{4}\left(\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\frac{1}{4}\left[\left(\frac{ab}{b+c}+\frac{ac}{b+c}\right)+\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\left(\frac{bc}{a+b}+\frac{ac}{a+b}\right)\right]\)
\(=\frac{1}{4}\left[\frac{a\left(b+c\right)}{b+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c\left(a+b\right)}{a+b}\right]\)
\(=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\cdot1=\frac{1}{4}\left(a+b+c=1\right)\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
trả lời
=1/3 nha
chúc bn
học tốt
