Problem
ALG-B1-M01-P017 Sum of Squares from Symmetry
#17
★★★☆☆ Level 3 of 5
Prove that \(a^2+b^2+c^2\ge ab+bc+ca\) for all real \(a,b,c\).
Move everything to one side and multiply by \(2\).
It is enough to prove \(a^2+b^2+c^2-ab-bc-ca\ge0\). Multiply by \(2\): \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \]
The right side is non-negative, so the inequality is proved.
This is not an inequalities module, but sums of squares are part of the language of identities.