Question

Cho a , b , c là các số dương . Chứng minh rằng :

\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{1}{2}\left(a+b+c\right)\)

đề thi học kì 2 đó giải giúp nhé ok

Answer 1

Ta có: \(\left(a+b\right)^2\ge4ab\)

\(\Rightarrow\dfrac{a+b}{4}\ge\dfrac{ab}{a+b}\)

\(\left(b+c\right)^2\ge4bc\)

\(\Rightarrow\dfrac{b+c}{4}\ge\dfrac{bc}{b+c}\)

\(\left(a+c\right)^2\ge4ac\)

\(\Rightarrow\dfrac{c+a}{4}\ge\dfrac{ac}{a+c}\)

Cộng từng vế BĐT, ta được:

\(\dfrac{a+b}{4}+\dfrac{b+c}{4}+\dfrac{a+c}{4}\ge\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ac}{a+c}\)

\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right)\ge\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ac}{a+c}\)

=> ĐPCM

Dấu = xảy ra khi: a = b = c