Can You Solve This Complex Quadratic Expression by Hand?

[anemone]

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

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

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

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

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 $$\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor=4.$$