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 $\sqrt 7  - \frac{m}{n} > \frac{1}{{mn}}$.

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 $7n^2 > m^2+3$.
