Can You Solve This Complex Quadratic Expression by Hand?

[anemone]

Here is this week's POTW:

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Evaluate \(\displaystyle \left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}\right\rfloor\) 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!

 

[anemone]

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 \(\displaystyle \left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor\) 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.

 

[anemone]

No one answered last week problem.:(

Here's my solution:

Spoiler

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

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

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

By the Cauchy-Schwarz inequality, we have:

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

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

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,

From $50>49$, we get: $\sqrt{2}>\dfrac{7}{5}$

From $12>9$, we get: $\sqrt{3}>\dfrac{3}{2}$

From $6>4$, we get: $\sqrt{6}>2$

Adding them up gives: $\sqrt{2}+\sqrt{3}+\sqrt{6}>4.9$

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

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