Evaluate [MATH]\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor[/MATH] without using a calculator.

My solution:

[MATH]\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)[/MATH]

[MATH]=\frac{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}{\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}[/MATH]

[MATH]=\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}[/MATH]

By the Cauchy-Schwarz inequality, we have:

[MATH]\begin{align*}\sqrt{2}+\sqrt{3}+\sqrt{6}&<\sqrt{1+1+1}\sqrt{2+3+6}\\&=\sqrt{33}\end{align*}[/MATH]

Hence [MATH]\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}>\frac{23}{\sqrt{33}}[/MATH].

From $528\lt 529$ we get, after taking the square root on both sides and rearranging:

$4\lt \dfrac{23}{\sqrt{33}}$

$\therefore \dfrac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\gt \dfrac{23}{\sqrt{33}}\gt 4$

On the other hand,

From $50\gt 49$, we get:

$\sqrt{2}\gt \dfrac{7}{5}$

From $12\gt 9$, we get:

$\sqrt{3}\gt \dfrac{3}{2}$

From $6\gt 4$, we get:

$\sqrt{6}\gt 2$

Adding them up gives:

$\sqrt{2}+\sqrt{3}+\sqrt{6}\gt 4.9$

$\therefore \dfrac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\lt \dfrac{23}{4.9}=4.69$.

We can conclude by now that [MATH]\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor=4.[/MATH]

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