Question

Cho x, y, z>0 và \(x+y+z\le1\). CM: \(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge10\)

Answer 1

Ta có : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)

Đặt \(Q=x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge x+y+z+\dfrac{9}{x+y+z}\)

\(=x+y+z+\dfrac{1}{x+y+z}+\dfrac{8}{x+y+z}\)

Áp dụng BĐT Cô - si có :

\(\left(x+y+z\right)+\dfrac{1}{x+y+z}\ge2\sqrt{\left(x+y+z\right)\cdot\dfrac{1}{x+y+z}}=2\)

Do \(x+y+z\le1\Rightarrow\dfrac{8}{x+y+z}\ge8\)

Do đó : \(Q\ge8+2=10\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Answer 2

\(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge x+y+z+\dfrac{9}{x+y+z}\)

\(VT\ge x+y+z+\dfrac{1}{x+y+z}+\dfrac{8}{x+y+z}\ge2\sqrt{\dfrac{x+y+z}{x+y+z}}+\dfrac{8}{1}=10\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)