2 −Fn+1 F2n

Problem 11339

(American Mathematical Monthly, Vol.115, January 2008) Proposed by J. L. D´ıaz-Barrero (Spain).

LetFnandLndenote thenth Fibonacci and Lucas numbers, respectively. Prove that for all n ≥ 1, 1

2

F_n^1/Fn+ L^1/L_n ⁿ

≤ 2 −Fn+1

F2n

.

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

It is easy to verify it for n = 1, 2, 3, 4:

n = 1 : 1 2

F1^1/F1+ L^1/L1 ¹

= 1 ≤ 1 = 2 −F2

F2

n = 2 : 1 2



F2^1/F2+ L^1/L2 ²



= 1 2

 1 +√³

3

≤ 4

3 = 2 − F3

F4

n = 3 : 1 2

F3^1/F3+ L^1/L3 ³

= 1 2

√ 2 +√⁴

4

≤15

8 = 2 −F4

F6

n = 4 : 1 2

F4^1/F4+ L¹4^/L4

= 1 2

√₃ 3 +√⁷

7

≤37

21 = 2 −F5

F8

Since x^1/x is a positive decreasing function for x ≥ e, and {Fn}n≥4 and {Ln}n≥4 are increasing sequences greater than e then

1 2

F_n^1/Fn+ L¹_n^/Ln is a decreasing sequence for n ≥ 4.

Moreover 2 − (Fn+1/F2n) is an increasing sequence for n ≥ 4 because 2 − (Fn+1/F2n) ≤ 2 − (Fn+2/F2n+2) that is equivalent to

(Fn+1+ Fn)F2n= Fn+2F2n ≤ Fn+1F2n+2= Fn+1(F2n+1+ F2n) or FnF2n ≤ Fⁿ⁺¹F2n+1which certainly holds since Fn is increasing. Therefore for n ≥ 4

1 2

F_n^1/Fn+ L^1/L_n ⁿ

≤1 2

F4^1/F4+ L^1/L4 ⁴

≤ 2 −F5

F8 ≤ 2 −Fn+1

F2n

.