Problem
ALG-B2-M06-P005 Schur form
#5
★★☆☆☆ Level 2 of 5
Prove that the inequality \(\sum a^3+3abc\ge\sum_{\mathrm{sym}}a^2b\) is equivalent to \[p^3-4pq+9r\ge0.\]
Hint. Substitute the formulas for \(\sum a^3\) and \(\sum_{\mathrm{sym}}a^2b\).
The difference between the left and right sides is \((p^3-3pq+3r)+3r-(pq-3r)=p^3-4pq+9r\). Therefore the original inequality is equivalent to the nonnegativity of this expression.
Connects the familiar Schur inequality with UVW language.