Condition that satisfies inequality involving fractions and the floor function

1. The floor function obeys$$\mathrm{ bn\le a<b(n+1),y=b, n\in\Bbb N}$$ Graph,
2. Using the identity above: $$\mathrm{y_1=b=\left\lfloor\frac ab\right\rfloor>y_2=\frac a{b+1}}$$
3. Therefore when we let a vary:$$\mathrm{a>b(b+1)=b^2+b, bn\le a<b(n+1),y=b, a,b,n\in\Bbb N}$$
4. Graphical proof

But this only gives the bound when the two graphs no longer cross and $a\ge \left\lfloor \frac ab \right \rfloor$. Modifying this a little bit after doing some pattern analysis of the inequalities satisfying b=1,2,... gives: $$\mathrm{a>n,b(n+1)\le a <(b+1)(n+1)=bn+b+n+1, a,b,n \in \Bbb N^0}$$

or simply: $$\mathrm{a>n,b(n+1)\le a <(b+1)(n+1)=bn+b+n+1, a,b,n \in \Bbb N^0, n\le b,\ if\ n=b,\ then \ a>b^2,\ or\ otherwise\ undefined}$$ Here is graphical proof of this one.

All in all:$$\mathrm{p>log_{10}(n)-7,0\le a-b(n+1)=10^{7+p}-bn-b<n+1⇔log_{10}(b)+log_{10}(n+1)-7 \le p < log_{10}(b+1)+log_{10}(n+1)-7, a,b,n\in \Bbb N^0=0,1,2,...,p\in\Bbb Z}$$

I will leave it to you to find other forms and edit my solutions.

These formulas even work with $a=x^x$ and other functions. Please correct me and give me feedback!

Let $a=bk+r$, where $0\leq r\leq {b-1}.$ Then we have, $\lfloor {\frac ab} \rfloor=k.$ So, $k>\frac {a}{b+1}.$ So that, $bk+k>bk+r.$ Which means, $k>r$. So, the range will be $a>b(r+1).$ So the range of $a$ for which your inequality is satisfied depends upon what $a$ mod $b$ is. But $a>b^2$ will invariably give safe values.