MHB Can You Solve This Complex Quadratic Expression by Hand?

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    2016
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The discussion centers around evaluating a complex quadratic expression without a calculator. The original problem contained an error regarding the sign of one term, which has since been corrected. The revised expression now reads: Evaluate the floor of the product of three square root terms. The author has extended the time for participants to solve the previous week's problem due to this oversight. An apology is issued for the mistake, emphasizing a commitment to accuracy in future problems.
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Here is this week's POTW:

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

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Hi all!

There is a glaring error about the sign of one of the terms that makes last week High School POTW unsolvable which I accidentally overlooked it.:(

It should read:

Evaluate $$\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor$$ without using a calculator.I will hence extend the period of time to solve for last week High School POTW for another 48 hours.

I want to apologize for making the mistake and I want to assure you that it will never happen again.
 
No one answered last week problem.:(

Here's my solution:

$$\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)$$

$$=\frac{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}{\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}$$

$$=\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}$$

By the Cauchy-Schwarz inequality, we have:

$$\begin{align*}\sqrt{2}+\sqrt{3}+\sqrt{6}&<\sqrt{1+1+1}\sqrt{2+3+6}\\&=\sqrt{33}\end{align*}$$

Hence $$\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}>\frac{23}{\sqrt{33}}$$.

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

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

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

On the other hand,

[TABLE="class: grid, width: 800"]
[TR]
[TD]From $50>49$, we get:

$\sqrt{2}>\dfrac{7}{5}$[/TD]
[TD]From $12>9$, we get:

$\sqrt{3}>\dfrac{3}{2}$[/TD]
[TD]From $6>4$, we get:

$\sqrt{6}>2$[/TD]
[TD]Adding them up gives:

$\sqrt{2}+\sqrt{3}+\sqrt{6}>4.9$[/TD]
[/TR]
[/TABLE]

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

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