Problem:
Let a,b and c be positive real numbers. Prove that
(a5−a2+3)(b5−b2+3)(c5−c2+3)≥(a+b+c)3
Solution:
For any positive number x, the quantities x2−1 and x3−1 have the same sign. Thus, we have 0≤(x3−1)(x2−1)=x5−x3−x2+1, or
x5−x2+3≥x3+2
It follows that
(a5−a2+3)(b5−b2+3)(c5−c2+3)≥(a3+2)(b3+2)(c3+2)
It suffices to show that
(a3+2)(b3+2)(c3+2)≥(a+b+c)3(*)
We finish with two approaches.
- First approach Expanding both sides of inequality (∗) and cancelling like terms gives a3b3c3+3(a3+b3+c3)+2(a3b3+b3c3+c3a3)+8≥3(a2b+b2a+b2c+c2b+c2a+ac2)+6abc.
By the AM-GM Inequality, we have a3+a3b3+1≥3a2b. Combining similar results, inequality (∗) reduces to
a3b3c3+a3+b3+c3+1+1≥6abc
which is evident by the AM-GM Inequality.
- We rewrite the left-hand-side of inequality (∗) as
(a3+1+1)(1+b3+1)(1+1+c3)
By Hölder's Inequality, we have
(a3+1+1)31(1+b3+1)31(1+1+c3)31≥(a+b+c)
from which inequality ( ∗ ) follows.
Problem originally by Titu Andreescu.
The problems on this page are the property of the MAA's American Mathematics Competitions