Question

Chứng minh BĐT :

\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{a+b+c}{2}\)(a,b,c>0)

Answer 1

ta có\(\frac{ab}{a+b}=\frac{4ab}{4\left(a+b\right)}=\frac{2ab+2ab}{4\left(a+b\right)}\le\frac{a^2+b^2+2ab}{4\left(a+b\right)}=\frac{\left(a+b\right)^2}{4\left(a+b\right)}=\frac{a+b}{4}\)

CMTT  ta được \(\frac{bc}{b+c}\le\frac{b+c}{4}và\frac{ca}{c+a}\le\frac{c+a}{4}\)

=>\(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a}\le\frac{a+b+b+c+c+a}{4}=\frac{2\left(a+b+c\right)}{4}=\frac{a+b+c}{2}\)