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.