Fibonacci inequality

Question

If $F_n$ denotes the $n$-th Fibonacci number ($F_0 = 0, F_1 = 1, F_{n+2} = F_{n+1} + F_n$), show that the inequalities

$F_{2n-2} < F_n^2 < F_{2n-1}$

hold for all $n ≥ 3$.

Answer 1

$F_{2(n-1)} = F_{n-1}\left(F_n + F_{n-2}\right) \to F_{n-1}\left(F_{n-1}+F_{n-2}+F_{n-2}\right) < \left(F_{n-1}+F_{n-2}\right)^2 \iff 0 < F_{n-2}^2$ which is clearly true.

$F_{2n-1} = F_{2n} - F_{2n-2} = F_n\left(F_{n+1} + F_{n-1}\right) - F_{n-1}\left(F_n + F_{n-2}\right) > F_n\left(F_n + 2F_{n-1}\right) - F_{n-1}\left(2F_n\right) = F_n^2$

Answer 2

We can use Binet's formula $$F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$$ where $$\alpha=\frac{1+\sqrt5}{2}\ ,\quad \beta=\frac{1-\sqrt5}{2}\ .$$ Therefore $$F_n^2>F_{2n-2}\quad\Leftrightarrow\quad (\alpha^n-\beta^n)^2>(\alpha-\beta)(\alpha^{2n-2}-\beta^{2n-2})\ .$$ It is easy to check that $\alpha\beta=-1$; rearranging the previous inequality therefore gives $$F_n^2>F_{2n-2}\quad\Leftrightarrow\quad (\alpha^{2n}-\alpha^{2n-1}-\alpha^{2n-3})+(\beta^{2n}-\beta^{2n-1}-\beta^{2n-3}) >2(-1)^n\ .$$ It is also easy to check that $\alpha^2=\alpha+1$ and hence that $\alpha^4-\alpha^3-\alpha=1$, and that the same holds for $\beta$, so the above becomes $$F_n^2>F_{2n-2}\quad\Leftrightarrow\quad \alpha^{2n-4}+\beta^{2n-4}>2(-1)^n\ .$$ If $n\ge3$ then $$LHS\ge\alpha^2=\frac{3+\sqrt5}{2}>2\ ,$$ which proves the inequality. The inequality $F_n^2<F_{2n-1}$ can be proved by the same methods (and in fact is very much easier).

Answer 3

We have $F_k$ = $F_{k-n}F_{n+1}+F_{k-n-1}F_n$ for k>n.

Thus, $F_{2n-2}<{F_n}^2\iff$$F_{n-2}F_{n+1}+F_{n-3}F_n<{F_n}^2\iff$$F_{n-2}F_{n+1}<{F_n}^2-F_{n-3}F_n\iff$$F_{n-2}F_{n+1}<{F_n}(2F_{n-2})\iff$$F_{n+1}<2F_n$ which is trivial.

Also, $F_{2n-1}>{F_n}^2\iff$$F_{n-1}F_{n+1}+F_{n-2}F_n>{F_n}^2\iff$$F_{n-1}F_{n+1}>F_{n-1}F_n$ which is again trivial.

Answer 4

$F_{ m+n }=F_{ n+1 }F_{ m }+F_{ n }F_{ m-1 }=F_{ n+1 }F_{ m+1 }-F_{ n-1 }F_{ m-1 }$ implies that

$F_{ 2n-1 }={ F }_{ n-1 }^{ 2 }+{ F }_{ n }^{ 2 }$

$F_{ 2n-2 }={ F }_{ n }^{ 2 }-{ F }_{ n-2 }^{ 2 }$

Therefore, $F_{ 2n-1 }={ F }_{ n-1 }^{ 2 }+{ F }_{ n }^{ 2 }>{ F }_{ n }^{ 2 }>{ F }_{ n }^{ 2 }-{ F }_{ n-2 }^{ 2 }=F_{ 2n-2 }$