Question

Cho a, b ,c >0. CM: \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Answer 1

* Áp dụng BĐT \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với $x,y>0$ vào bài toán có :

\(\dfrac{1}{4}\cdot\left(\dfrac{4}{a+b}\right)\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{b+c}\right)\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{c+a}\right)\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế với vế các BĐT có :

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

Dấu "=" xảy ra khi \(a=b=c\)