Olympiad Maths Prep

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Problem 993

AIME late
Algebra Difficulty 6.0 Prove it

8-178 The sequence {Fn}\left\{F_{n}\right\} is defined as follows: F1=1,F2=2F_{1}=1, F_{2}=2, and
Fn+2=Fn+1+Fn,n=1,2,3,⋯ F_{n+2}=F_{n+1}+F_{n}, n=1,2,3, \cdots

Prove that for any natural number nn, we have
Fn+1n⩾1+1Fnn. \sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}} .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solutions — 2

Solution 1

[Proof] Let F0=1F_{0}=1, then
Fk+1=Fk+Fk−1,  F_{k+1}=F_{k}+F_{k-1} \text {, }

i.e., 1=FkFk+1+Fk−1Fk+1,k=1,2,3,⋯1=\frac{F_{k}}{F_{k+1}}+\frac{F_{k-1}}{F_{k+1}}, k=1,2,3, \cdots
Thus,
n=∑k=1nFkFk+1+∑k=1nFk−1Fk+1. n=\sum_{k=1}^{n} \frac{F_{k}}{F_{k+1}}+\sum_{k=1}^{n} \frac{F_{k-1}}{F_{k+1}} .

By the AM-GM inequality, we have
1⩾F1F2⋅F2F3⋯⋅FnFn+1n+F0F2⋅F1F5⋅F2F4⋯⋅Fn−1Fn+1n. 1 \geqslant \sqrt[n]{\frac{F_{1}}{F_{2}} \cdot \frac{F_{2}}{F_{3}} \cdots \cdot \frac{F_{n}}{F_{n+1}}}+\sqrt[n]{\frac{F_{0}}{F_{2}} \cdot \frac{F_{1}}{F_{5}} \cdot \frac{F_{2}}{F_{4}} \cdots \cdot \frac{F_{n-1}}{F_{n+1}}} .

Noting that F0=F1=1F_{0}=F_{1}=1, we get
1⩾1Fn+1n+1FnFn+1n, 1 \geqslant \frac{1}{\sqrt[n]{F_{n+1}}}+\frac{1}{\sqrt[n]{F_{n} F_{n+1}}},

i.e., Fn+1n⩾1+1Fnn\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}}.

Solution 2

[Proof] Let F0=1F_{0}=1, then
Fk+1=Fk+Fk−1F_{k+1}=F_{k}+F_{k-1}

i.e. □\square
1=FkFk+1+Fk−1Fk+1,k=1,2,3,⋯1=\frac{F_{k}}{F_{k+1}}+\frac{F_{k-1}}{F_{k+1}}, k=1,2,3, \cdots

Thus,
n=∑k=1nFkFk+1+∑k=1nFk−1Fk+1n=\sum_{k=1}^{n} \frac{F_{k}}{F_{k+1}}+\sum_{k=1}^{n} \frac{F_{k-1}}{F_{k+1}}

By the AM-GM inequality, we have
1⩾F1F2⋅F2F3⋯⋅FnFn+1n+F0F2⋅F1F5⋅F2F4⋯⋅Fn−1Fn+1n1 \geqslant \sqrt[n]{\frac{F_{1}}{F_{2}} \cdot \frac{F_{2}}{F_{3}} \cdots \cdot \frac{F_{n}}{F_{n+1}}}+\sqrt[n]{\frac{F_{0}}{F_{2}} \cdot \frac{F_{1}}{F_{5}} \cdot \frac{F_{2}}{F_{4}} \cdots \cdot \frac{F_{n-1}}{F_{n+1}}}

Noting that F0=F1=1F_{0}=F_{1}=1, we get
1⩾1Fn+1n+1FnFn+1n1 \geqslant \frac{1}{\sqrt[n]{F_{n+1}}}+\frac{1}{\sqrt[n]{F_{n} F_{n+1}}}

i.e. Fn+1n⩾1+1Fnn\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.