Problem
ALG-B1-M07-P010 Fixed product
#10
★★☆☆☆ Level 2 of 5
Let \(x,y,z>0\) and \(xyz=1\). Prove that \(x+y+z\ge3\).
Apply AM-GM to \(x,y,z\).
By AM-GM, \(\frac{x+y+z}{3}\ge\sqrt[3]{xyz}=1\). Therefore \(x+y+z\ge3\). Equality occurs when \(x=y=z=1\).
A basic transition from product to sum.