Methods of Proof 327The induction is complete.
13.As in the solution to the previous problem we argue by induction onnusing trigono-
metric identities. The base case holds because
|sinx 1 |+|cosx 1 |≥sin^2 x 1 +cos^2 x 1 = 1.Next, assume that the inequality holds forn=kand let us prove it forn=k+1. Using
the inductive hypothesis, it suffices to show that
|sinxn+ 1 |+|cos(x 1 +x 2 +···+xn+ 1 )|≥|cos(x 1 +x 2 +···+xn)|.To simplify notation letxn+ 1 =xandx 1 +x 2 +···+xn+xn+ 1 =y, so that the inequality
to be proved is|sinx|+|cosy|≥|cos(y−x)|. The subtraction formula gives
|cos(y−x)|=|cosycosx+sinysinx|≤|cosy||cosx|+|siny||sinx|
≤|cosy|+|sinx|.This completes the inductive step, and concludes the solution.
(Revista Mathematica din Timi ̧soara(Timi ̧soara Mathematics Gazette), proposed by
T. Andreescu)
14.We expect an inductive argument, with a possible inductive step given by
3 n+^1 = 3 · 3 n≥ 3 n^3 ≥(n+ 1 )^3.In order for this to work, the inequality 3n^3 ≥(n+ 1 )^3 needs to be true. This inequality
is equivalent to 2n^3 ≥ 3 n^2 + 3 n+1, which would, for example, follow from the separate
inequalitiesn^3 ≥ 3 n^2 andn^3 ≥ 3 n+1. These are both true forn≥3. Thus we can
argue by induction starting with the base casen=3, where equality holds. The cases
n=0,n=1, andn=2 can be checked by hand.
15.The base case 2^6 < 6 !< 36 reduces to 64< 720 <729, which is true. Assuming
the double inequality true fornwe are to show that
(
n+ 1
3)n+ 1
<(n+ 1 )!<(
n+ 1
2)n+ 1
.Using the inductive hypothesis we can reduce the inequality on the left to
(
n+ 1
3)n+ 1
<(n+ 1 )(n
3)n
,or