Ta có bđt \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)(1)
Chứng minh:
Áp dụng bđt cosi cho 3 số dương:
\(x+y+z\ge3\sqrt[3]{xyz}\left(2\right)\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge3\sqrt[3]{\dfrac{1}{xyz}}\)(3)
Từ (2),(3)\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\dfrac{1}{xyz}}=9\)
Vậy bđt (1) đã chứng minh
Áp dụng bđt (1), ta có \(\left[\left(2a+b\right)+\left(2b+c\right)+\left(2c+a\right)\right]\left(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\right)\ge9\Leftrightarrow3\left(a+b+c\right)\left(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\right)\ge9\Leftrightarrow3.1.\left(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\right)\ge9\Leftrightarrow\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\ge3\)Vậy nếu a+b+c=1 thì \(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\ge3\)
