Question

Cho x,y,z>0 và \(xyz\ge x+y+z+2\). Tìm Min(x+y+z)

Answer 1

1;2;3

Answer 2

Ta có:

\(xyz\ge x+y+z+2\ge2+3\sqrt[3]{xyz}\)

\(\Leftrightarrow\frac{x+y+z}{3}\ge\sqrt[3]{xyz}\ge2\)

\(\Leftrightarrow x+y+z\ge6\)

Answer 3

\(xyz\ge x+y+z+2.\)

Ta có :

\(xyz\ge x+y+z+2\ge2+3\sqrt[3]{xyz}\)

\(\Leftrightarrow\frac{x+y+z}{3}\ge\sqrt[3]{xyz}\ge2\)

\(\Leftrightarrow x+y+z\ge6\)