Next note that if we get $\gcd(m,n)\cdot \gcd(m+1,n+1) \leq m-n$, we are done.
To prove it, note that $d_1 = \gcd(m,n) = \gcd(n,m-n)$ which divides $m-n$, and $d_2=\gcd(m+1,n+1)=\gcd(n+1,m-n)$ which also divides $m-n$. But $d_1$ divides $n$ and $d_2$ divides $n+1$ ... and $\gcd(n,n+1)=1$ so in fact $\gcd(d_1,d_2)=1$ !.
Hence $d_1\cdot d_2$ must divide $m-n$, and so $d_1\cdot d_2 \leq m - n$ completing the proof $\square$.