$5775$ and $5776$ are two consecutive abundant numbers, i.e., $n$ and $n+1$ such that both $\sigma(n) > 2n$ and $\sigma(n+1) > 2(n+1)$ (see A096399).
Are there any two consecutive numbers $n$ and $(n+1)$ such that both $\sigma(n) > 3n$ and $\sigma(n+1) > 3(n+1)$?
You want numbers such that $\frac{\sigma(n)}{n}>3$.
Recall that if $n=\prod\limits_{i=1}^s p_i^{n_i}$ then $\frac{\sigma(n)}{n}=\prod_{i=1}^s \frac{1-(1/p_i)^{n_i+1}}{1-(1/p)}$.
Now, it is not hard to show that $\prod\limits_{i=1}^\infty\frac{1-(1/p_i)^2}{1-(1/p)}=\prod\limits_{i=1}^\infty 1+\frac{1}{p}$ diverges.
This means that we can find distinct primes $a_1,a_2\dots a_m$ and $b_1,b_2\dots b_l$ so that $\prod\limits_{i=1}^m 1+\frac{1}{a_i}>3$ and $\prod\limits_{i=1}^m 1+\frac{1}{b_i}>3$.
It is a direct consequence of the chinese remainder theorem that there exists and integer $n$ such that $n\equiv 0 \bmod a_i$ and $n\equiv -1 \bmod b_j$ for all suitable $i,j$.
Therefore we must have that $\frac{\sigma(n)}{n}>3$ and $\frac{\sigma(n+1)}{n+1}>3$ as desired.