# 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]

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.
• $\beta = (7b^2 - n^2)/(b\sqrt{7} + n) \geq 1/(b\sqrt{7} + n) = 1/(n + \beta + n) = 1/(2n + \beta)$. Thus $2n + \beta \geq 1/\beta$. – user187373 Oct 31 '14 at 6:04
• @MarkHurd I fell for that too for a second. The role that $2$ plays in $\alpha\geq\frac{1}{2m+\alpha}$ is not coming from $\sqrt{2}$. It comes from replacing $a\sqrt{2}+m$ with $(m+\alpha)+m$. And so the same thing happens with $\beta$: $b\sqrt{7}+n\to(n+\beta)+n$. – alex.jordan Oct 31 '14 at 6:05
• So, using this method we can show for any non-integers $\sqrt{m}$ and $\sqrt{n}$, right? – simmons Oct 31 '14 at 20:22