Maths Olympiad Prep

Library / /2 of 45

Algebra Difficulty 4.7 AIME Prove it Romania

Suppose a∈[−2,∞)a \in [-2, \infty), r∈[0,∞)r \in [0, \infty) and let nn be a positive integer.
Show that
r2n+arn+1≥(1−r)2n. r^{2n} + a r^n + 1 \geq (1 - r)^{2n}.

Solution

If r=0r = 0, the relation is obvious.
Otherwise, dividing by r2nr^{2n}, one gets the same inequality with rr replaced by 1r\frac{1}{r}, so one can assume r∈(0,1]r \in (0, 1].
Since r2n+arn+1≥r2n−2rn+1=(1−rn)2r^{2n} + a r^n + 1 \geq r^{2n} - 2 r^n + 1 = (1 - r^n)^2 and 1−rn≥01 - r^n \geq 0, it is enough to prove that 1−rn≥(1−r)n1 - r^n \geq (1 - r)^n.
This follows from rn+(1−r)n≤r+(1−r)=1r^n + (1 - r)^n \leq r + (1 - r) = 1.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.