Question

[thyth]

Cho a,b,c >0.CMR:

\(\dfrac{1}{2\cdot a+b}+\dfrac{1}{2\cdot b+c}+\dfrac{1}{2\cdot c+a}>=\dfrac{3}{a+b+c}\)

Answer 1

[Mysterious Person]

áp dụng cô si ta có :

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

\(=\dfrac{9}{3\left(a+b+c\right)}=\dfrac{3}{a+b+c}\)