Problem
ALG-B2-M11-P018 Squares and cubes
#18
★★★★★ Level 5 of 5
Prove for \(a,b,c\ge0\): \[(a^2+b^2+c^2)^3\ge(a^3+b^3+c^3)^2.\]
Hint. First bound \((\sum a^3)^2\) using \(\sum a^2\) and \(\sum a^4\).
By Cauchy, \[(\sum a^3)^2=(\sum a\cdot a^2)^2\le(\sum a^2)(\sum a^4).\] Also, \(\sum a^4\le(\sum a^2)^2\). Therefore \((\sum a^3)^2\le(\sum a^2)^3\).
Teaching goal: the student should first recognise the type of estimate, then choose the tool.