The inequality (1) is equivalent to Fk−1Fk+2&lt

Problem 12213

(American Mathematical Monthly, Vol.127, November 2020) Proposed by H. Ohtsuka (Japan).

For n ≥ 1, prove _n

X

k=1

pFk

−1F_k+2≤p

F_n+1F_n+4−√ 5

where Fn is the n-th Fibonacci number.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. For n = 1 the equality holds. For n ≥ 2, we show that the strict inequality holds

n

X

k=1

pFk−1F_k+2<p

F_n+1F_n+4−√ 5.

It suffices to show that for k ≥ 2, pFk

−1F_k+2<p

F_k+1F_k+4−p

FkF_k+3 (1)

then

n

X

k=1

pFk

−1F_k+2=

n

X

k=2

pFk

−1F_k+2<

n

X

k=2

p

F_k+1F_k+4−p

FkF_k+3

=p

F_n+1F_n+4−p F₂F₅

and the given inequality follows because F0F₃= 0 and F2F₅= 5.

The inequality (1) is equivalent to

Fk−1F_k+2< F_k+1F_k+4+ F^kF_k+3− 2p

FkF_k+1F_k+3F_k+4 that is

2p

F_kF_k+1F_k+3F_k+4< F_k+1F_k+4+ F^kF_k+3− F^k−1F_k+2

= (Fk+2− F^k)(Fk+2+ Fk+3) + F^kF_k+3− F^k−1F_k+2

= Fk+2(Fk+2+ Fk+3− F^k− F^k−1) = 2Fk+2² . Hence it remains to show that

FkF_k+1F_k+3F_k+4< F_k+2⁴

which holds, because by Cassini’s identity Fⁿ₋^rF_n+r = Fn²− (−1)^n−rF_r², FkF_k+1F_k+3F_k+4= (F^kF_k+4)(Fk+1F_k+3)

= (Fk+2² − (−1)^k)(Fk+2² + (−1)^k)

= Fk+2⁴ − 1 < Fk+2⁴ .