Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ta có:
\(\dfrac{ab}{c+1}=\dfrac{ab}{\left(c+a\right)+\left(b+c\right)}\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại
\(\dfrac{bc}{a+1}\le\dfrac{1}{4}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{b+1}\le\dfrac{1}{4}\left(\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{ab+bc}{a+c}+\dfrac{bc+ac}{a+b}+\dfrac{ab+ac}{b+c}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{b+c}\right)\)
\(=\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\cdot1=\dfrac{1}{4}\left(a+b+c=1\right)\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)
