Problem
ALG-B2-M12-P011 Set 3. Holder in a cycle
#11
★★★★★ Level 5 of 5
Prove for \(a,b,c>0\): \[\sum_{\mathrm{cyc}}\frac{a^3}{b^2+bc+c^2}\ge\frac{a+b+c}{3}.\]
Hint. Use Holder.
By Holder, the left side is at least \(\frac{(a+b+c)^3}{2(a^2+b^2+c^2)+ab+bc+ca}\). Since the denominator is at most \(3(a+b+c)^2\), the result follows.
Mock set 3: the problem is intended for independent method selection without an explicit cue in the statement.