Problem
ALG-B2-M01-P001 Sum of three squares
#1
★★☆☆☆ Level 2 of 5
Prove that for all real \(a,b,c\), \(a^2+b^2+c^2\ge ab+bc+ca\).
Hint 1. Multiply the difference by \(2\).
Hint 2. Try to obtain a sum of squares of differences.
We have \[2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\ge0.\] Hence the inequality is true. Equality occurs exactly when \(a=b=c\).
Module training problem. Method tags: squares, equality-case.