For positive integers $m,n$ if $\sqrt 7 - \frac{m}{n} > 0$ then prove that $\sqrt 7 - \frac{m}{n} > \frac{1}{{mn}}$ For positive integers $m,n$ if $\sqrt 7  - \frac{m}{n} > 0$ then prove that $\sqrt 7  - \frac{m}{n} > \frac{1}{{mn}}$.
 A: By condition, $7n^2 > m^2$. The square of an integer when divided by 7 moiety can be given in only 0, 1, 2 and 4. Therefore, none of the numbers $m^2+1$, $m^2+2$ is not divisible by 7, where $7n^2\ge m^2+3$. Then $n\sqrt7\ge\sqrt{m^2+3}\ge\sqrt{m^2+2+\frac1{m^2}} > m+\frac1m$ at $m > 1$, so $\sqrt7-\frac{m}n > \frac1{mn}$.
Case $m=1$ immediately obvious, but you can also notice that in this case, true strict inequality $7n^2 > m^2+3$.
A: I have tried like this.
First let, $\sqrt 7 n = i + f$ where $i = \left\lfloor {\sqrt 7 n} \right\rfloor $ and $f = \left\langle {\sqrt 7 n} \right\rangle $.
Let $m = i - p$ where $p \in \mathbb{Z}$ such that $1 \leqslant m < \sqrt 7 n$.
So we have to prove, $\sqrt 7 mn > {m^2} + 1$, i.e. $\left( {i + f} \right)\left( {i - p} \right) > {\left( {i - p} \right)^2} + 1 \Rightarrow \left( {i - p} \right)\left( {f + p} \right) > 1$ which is obviously true for $p\ge1$. For $p=0$ and $n\ge 4$ ,$\left( {i - p} \right)\left( {f + p} \right) > 1$ because in this case $\left\lfloor {\sqrt 7 n} \right\rfloor  \geqslant 10$. For $p=0$ and $0<n<4$ we have to simply check the inequality.
