[sp]Suppose that there exist $m,n$ such that $0 < \sqrt7 - \frac mn < \frac1{mn}$. Then $\frac mn < \sqrt7 < \frac{m^2+1}{mn}$, and (after squaring and multiplying through by $m^2n^2$) $$ m^4 < 7m^2n^2 < (m^2+1)^2 = m^4 + 2m^2 + 1,$$ $$0 < m^2(7n^2-m^2) < 2m^2+1,$$ $$ 0 < 7n^2 - m^2 < 2 + \tfrac1{m^2}.$$ Since $7n^2 - m^2$ is an integer, it must therefore be $1$ or $2$. But the square of an integer is congruent to $0$ or $1\pmod4$, so $7n^2-m^2 = 1\pmod4$ can never occur; and the only possible solution to $7n^2-m^2 = 2\pmod4$ is if $m$ and $n$ are both odd. But the square of an odd number is congruent to $1\pmod8$. So if $m$ and $n$ are both odd then $7n^2-m^2 = 6\ne2\pmod8$. Therefore there are no solutions. It follows that if $\sqrt 7 - \frac {m}{n}$ is greater than $0$ then it must be greater than $\frac1{mn}.$Albert said:$m,n\in N$ , and $\sqrt 7 - \dfrac {m}{n}>0$
prove :
$\sqrt 7 - \dfrac {m}{n}>\dfrac {1}{m\times n}$
nice solution !Opalg said:[sp]Suppose that there exist $m,n$ such that $0 < \sqrt7 - \frac mn < \frac1{mn}$. Then $\frac mn < \sqrt7 < \frac{m^2+1}{mn}$, and (after squaring and multiplying through by $m^2n^2$) $$ m^4 < 7m^2n^2 < (m^2+1)^2 = m^4 + 2m^2 + 1,$$ $$0 < m^2(7n^2-m^2) < 2m^2+1,$$ $$ 0 < 7n^2 - m^2 < 2 + \tfrac1{m^2}.$$ Since $7n^2 - m^2$ is an integer, it must therefore be $1$ or $2$. But the square of an integer is congruent to $0$ or $1\pmod4$, so $7n^2-m^2 = 1\pmod4$ can never occur; and the only possible solution to $7n^2-m^2 = 2\pmod4$ is if $m$ and $n$ are both odd. But the square of an odd number is congruent to $1\pmod8$. So if $m$ and $n$ are both odd then $7n^2-m^2 = 6\ne2\pmod8$. Therefore there are no solutions. It follows that if $\sqrt 7 - \frac {m}{n}$ is greater than $0$ then it must be greater than $\frac1{mn}.$
[Notice that if you drop the condition $\sqrt 7 - \frac {m}{n}>0$ then it is possible to have $\Bigl|\sqrt 7 - \frac {m}{n}\Bigr| < \frac1{mn}.$ For example, if $m=8$ and $n=3$ then $\Bigl|\sqrt 7 - \frac83\Bigr| \approx 0.0209 < \frac1{24} \approx 0.0417.$ But in that case, $\sqrt 7 - \frac 83$ is negative.][/sp]
This is an inequality that is asking for a proof that the expression 7^0.5 - (m/n) is greater than 1/(mn).
To prove this inequality, you can use algebraic manipulations and properties of exponents to simplify the expression and compare the values.
There are various methods and approaches that can be used to prove an inequality. Some common ones include using algebraic manipulations, mathematical induction, and contradiction.
Yes, this inequality can be proven for all values of m and n as long as they are non-zero real numbers.
This inequality may have various applications in different mathematical and scientific fields, such as in optimization problems or in the study of exponential functions and their properties. It may also be used as a tool to prove other related inequalities.