Prove that if $\sigma(F_a)<\sigma(F_b)$, where $1 \leq b < a$, then $\sigma(a)<\sigma(b)$ also. Here $\sigma(x)$ denotes the sum of the divisors of $x$ and $F_x$ is the $x$th Fibonacci number. Additionally, a highly abundant number is a number $k$ for which $\sigma(k) > \sigma(m) \hspace{3 mm} \forall m < k$.
As a special case of the above, if we define $a(n)=\sigma(F_n)$, then whenever $a(n)$ is not a record value of the sequence, n is not highly abundant.
The (now deleted) second part of this question has been moved here so that this one, which has been disproved by @Greg Martin, may be closed.
The assertion is false for $a=82$ and $b=81$. The 81st and 82nd Fibonacci numbers are $37889062373143906 = 2\times17\times53\times109\times2269\times4373\times19441$ and $61305790721611591 = 2789\times59369\times370248451$ respectively. So \begin{align*} \sigma(37889062373143906) &= 3\times18\times54\times110\times2270\times4374\times19442 \\ &= 61919263145361600, \text{ which is larger than} \\ \sigma(61305790721611591) &= 2790\times59370\times370248452 \\ &= 61328805160719600; \end{align*} but $\sigma(81) = 121$ while $\sigma(82)=126$.