Exercise 191Algebra › InequalitiesHigh schoolAdded 31 Oct 2026

Squares and pairwise products

#inequalities#squares

Statement

Prove that for all real numbers $a,b,c$ ,
$a^2+b^2+c^2\ge ab+bc+ca,$
with equality if and only if $a=b=c$ .

Hint

Look at $(a-b)^2+(b-c)^2+(c-a)^2$ .

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Expanding each square,
$(a-b)^2+(b-c)^2+(c-a)^2=2\left(a^2+b^2+c^2-ab-bc-ca\right).$

A sum of squares is non-negative, so $a^2+b^2+c^2-ab-bc-ca\ge0$ . It equals $0$ exactly when all three squares are $0$ , that is when $a=b=c$ .