Problem
ALG-B2-M04-P018 Fifth power and product of sums
#18
★★★★★ Level 5 of 5
Prove for \(a,b,c\ge0\): \[3(a^5+b^5+c^5)\ge(a^2+b^2+c^2)(a^3+b^3+c^3).\]
Hint. Order the numbers. The sequences \(a^2,b^2,c^2\) and \(a^3,b^3,c^3\) have the same order.
After ordering, \(a^2,b^2,c^2\) and \(a^3,b^3,c^3\) increase together. By Chebyshev, \[\frac{a^5+b^5+c^5}{3}\ge\frac{a^2+b^2+c^2}{3}\cdot\frac{a^3+b^3+c^3}{3}.\] Multiplying by \(9\) gives the result.
The student must choose the two powers, not only \(a\) and \(a^2\).