Floor function inequality $\lfloor a\sqrt{2}\rfloor\lfloor b\sqrt{7}\rfloor <\lfloor ab\sqrt{14}\rfloor$ Let $a,b$ be positive integers. Show that $\lfloor a\sqrt{2}\rfloor\lfloor b\sqrt{7}\rfloor <\lfloor ab\sqrt{14}\rfloor$.
[Source: Russian competition problem]
 A: Let $a\sqrt{2} = m + \alpha$ where $m$ is an integer and $0 < \alpha < 1$, and $b\sqrt{7} = n + \beta$ where $n$ is an integer and $0 < \beta < 1$. (The fractional parts are positive since $\sqrt{2}$ and $\sqrt{7}$ are irrational.) We are to prove that
$$ (m+ \alpha)(n + \beta) - mn \geq 1.$$
We have 
$$\alpha = a\sqrt{2} - m = \frac{2a^2 - m^2}{a\sqrt{2} + m} = \frac{2a^2 - m^2}{2m + \alpha} \geq \frac{1}{2m + \alpha},$$
since $2a^2 - m^2$ is necessarily a positive integer. From this it is easy to see that 
$$m \geq \frac{1}{2}\left(\frac{1}{\alpha} - \alpha\right).$$ 
We similarly prove $\beta \geq 1/(2n + \beta)$, hence
$$n \geq \frac{1}{2}\left(\frac{1}{\beta} - \beta\right).$$
Now we have
$$
\begin{align*}
(m + \alpha)(n + \beta) - mn &= \alpha n + \beta m + \alpha \beta \\
&\geq \frac{\alpha}{2}\left(\frac{1}{\beta} - \beta\right) + \frac{\beta}{2}\left(\frac{1}{\alpha} - \alpha\right) + \alpha \beta \\
&=\frac{1}{2}\left( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \right)\\
&= 1 + \frac{(\alpha - \beta)^2}{2\alpha\beta}\\
&\geq 1,
\end{align*}
$$
as required.
