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Algebra Difficulty 5.6 AIME, harder Prove it

6. The sequence {Fn}\left\{F_{n}\right\} is defined as follows: F1=1,F2=2F_{1}=1, F_{2}=2, and for any n∈Nn \in N, Fn+2=Fn+1+FnF_{n+2}=F_{n+1}+F_{n}. Prove: For any n∈Nn \in N, Fn+1n⩾1+1Fnn\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}}.

Solution

6. Let F0=Fˉ2−F1=1F_{0}=\bar{F}_{2}-F_{1}=1, then
Fk+1=Fk+Fk−1,k=1,2,⋯ ,n,  F_{k+1}=F_{k}+F_{k-1}, k=1,2, \cdots, n \text {, }

which means
1=FkFk+1+Fk−1Fk+1,k=1,2,⋯ ,n. 1=\frac{F_{k}}{F_{k+1}}+\frac{F_{k-1}}{F_{k+1}}, k=1,2, \cdots, n .

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}}, hence
1=1n∑k=1nFkFk+1+1n∑k=1nFk−1Fk+1⩾F1F2⋅F2F3⋅⋯⋅FnFn+1n+F0F2⋅F1F3⋅F2F4⋅⋯⋅Fn−1Fn+1n=1Fn+1n+1FnFn+1n,  \begin{array}{l} 1=\frac{1}{n} \sum_{k=1}^{n} \frac{F_{k}}{F_{k+1}}+\frac{1}{n} \sum_{k=1}^{n} \frac{F_{k-1}}{F_{k+1}} \\ \geqslant \sqrt[n]{\frac{F_{1}}{F_{2}} \cdot \frac{F_{2}}{F_{3}} \cdot \cdots \cdot \frac{F_{n}}{F_{n+1}}} \\ +\sqrt[n]{\frac{F_{0}}{F_{2}} \cdot \frac{F_{1}}{F_{3}} \cdot \frac{F_{2}}{F_{4}} \cdot \cdots \cdot \frac{F_{n-1}}{F_{n+1}}} \\ =\frac{1}{\sqrt[n]{F_{n+1}}}+\frac{1}{\sqrt[n]{F_{n} F_{n+1}}} \text {, } \\ \end{array}

which implies Fn+1n⩾1+1Fn\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F}}.

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