Question

cho a+b+c=1.chung minh:\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ac}{b+1}\le\frac{1}{4}\)

Answer 1

\(\frac{a+b}{ab}\ge\frac{4}{a+b}\)=>\(a+b\ge\frac{4ab}{a+b}\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)

=>\(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)

=>\(\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ac}{b+1}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{a+c}+\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)

                                             =\(\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)

dau bang xay ra <=>a=b=c=\(\frac{1}{3}\)