Problem:
Let a,b,c be real numbers greater than or equal to 1 . Prove that
min(b2−5b+1010a2−5a+1,c2−5c+1010b2−5b+1,a2−5a+1010c2−5c+1)≤abc
Solution:
We start by observing that the denominators of the fractions involved in the statement of the problem are positive. Next, we argue by contradiction and assume that
10a2−5a+1>abc(b2−5b+10)
and similar inequalities obtained by cyclic permutations. Multiplying these inequalities yields
∏[a3(a2−5a+10)]<∏(10a2−5a+1)
This is impossible, since
a3(a2−5a+10)−(10a2−5a+1)=(a−1)5≥0
and similarly for b and c.
This problem and solution was suggested by Titu Andreescu.
The problems on this page are the property of the MAA's American Mathematics Competitions