Problem
ALG-B1-M07-P001 Square of a difference
#1
★☆☆☆☆ Level 1 of 5
Prove that for all real \(a,b\), \(a^2+b^2\ge2ab\).
Start with \((a-b)^2\ge0\).
We have \((a-b)^2=a^2-2ab+b^2\ge0\). Hence \(a^2+b^2\ge2ab\). Equality occurs when \(a=b\).
Basic technique: proof through a nonnegative square.