Problem
ALG-B2-M10-P003 Squares versus abc
#3
★★★★★ Level 5 of 5
Prove for \(a,b,c\ge0\): \[(a^2+b^2+c^2)^2\ge3abc(a+b+c).\]
Hint. Use two estimates: \(\sum a^2\ge ab+bc+ca\) and \((ab+bc+ca)^2\ge3abc(a+b+c)\).
Let \(q=ab+bc+ca\). Then \(\sum a^2\ge q\), so \((\sum a^2)^2\ge q^2\). Applying \((x+y+z)^2\ge3(xy+yz+zx)\) to \(x=ab\), \(y=bc\), \(z=ca\), gives \(q^2\ge3abc(a+b+c)\).
Teaching goal: choose a combination of methods and always check the equality case.