Problem
ALG-B2-M06-P006 Squares versus products
#6
★★★☆☆ Level 3 of 5
Prove for \(a,b,c\ge0\): \[a^2+b^2+c^2\ge ab+bc+ca.\]
Hint. Use the formula for \(p^2-3q\).
The inequality is equivalent to \(p^2-2q\ge q\), i.e. \(p^2-3q\ge0\). But \(p^2-3q=\frac12((a-b)^2+(b-c)^2+(c-a)^2)\ge0\).
Shows how a basic inequality looks in \(p,q\)-form.