square root and inequality: prove that sqrt(6)+sqrt(12)+sqrt(21) < 11

[SimonAlpar]

Hello!
I need to proof (without computing the square roots!) that "sqrt(6)+sqrt(12)+sqrt(21) < 11". How to do it step by step?

 

[blamocur]

Hello!
I need to proof (without computing the square roots!) that "sqrt(6)+sqrt(12)+sqrt(21) < 11". How to do it step by step?

What have you tried? What subject is this homework related to?

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[Dr.Peterson]

Hello!
I need to proof (without computing the square roots!) that "sqrt(6)+sqrt(12)+sqrt(21) < 11". How to do it step by step?

Can you find an integer that is greater than [imath]\sqrt{6}[/imath]? [Hint: Find a square that is greater than 6.]

 

[SimonAlpar]

Can you find an integer that is greater than [imath]\sqrt{6}[/imath]? [Hint: Find a square that is greater than 6.]

yes, it is 3, 4 is greater than sqrt(12) and 5 than sqrt(21), but their sum is 12

 

[blamocur]

Can you find an integer that is greater than [imath]\sqrt{6}[/imath]? [Hint: Find a square that is greater than 6.]

You can get even better upper boundaries if you consider simple fractions.

 

[SimonAlpar]

You can get even better upper boundaries if you consider simple fractions.

thank you. it's ok now

 

[SimonAlpar]

great help!

 

[Dr.Peterson]

You can get even better upper boundaries if you consider simple fractions.

True; but that isn't required here. (And I was primarily giving an example of what can be done, which could be generalized if needed.)

 

[blamocur]

True; but that isn't required here. (And I was primarily giving an example of what can be done, which could be generalized if needed.)

Unless I am missing something limiting those roots with integers is not enough for the proof. But I agree about the direction.

 

[Dr.Peterson]

Unless I am missing something limiting those roots with integers is not enough for the proof. But I agree about the direction.

Actually, I misread the 21 as another 12. I still need to get new glasses.

 

[lookagain]

Hello!
I need to proof (without computing the square roots!) that "sqrt(6)+sqrt(12)+sqrt(21) < 11". How to do it step by step?

Estimation would still be involved if you were to go with this type of route/method (here is an idea):

Spoiler

\(\displaystyle \ \ \ \ \ \ \sqrt{6} \ < \ \ \sqrt{6.25} \ = \ ?\)
\(\displaystyle \ \ \ \ \sqrt{12} \ < \sqrt{12.25} \ = \ ?\)
\(\displaystyle + \ \sqrt{21} \ < \ \ \sqrt{21.16} \ = \ ?\)
- - - - - - - - - - - - - - - - -

_________________________________________________________________

Or, if you were to do it this way without estimating at all, but with multiple squaring and much
recombining needed (just two beginning steps are shown):

Spoiler

\(\displaystyle \sqrt{6} + \sqrt{12} \ \ \ vs. \ \ \ 11 - \sqrt{21}\)

\(\displaystyle (\sqrt{6} + \sqrt{12})^2 \ \ \ vs. \ \ \ (11 - \sqrt{21})^2\)


Continue on with about 10 steps, plus or minus a few steps.

 

[blamocur]

Estimation would still be involved if you were to go with this type of route/method (here is an idea):

Spoiler

\(\displaystyle \ \ \ \ \ \ \sqrt{6} \ < \ \ \sqrt{6.25} \ = \ ?\)
\(\displaystyle \ \ \ \ \sqrt{12} \ < \sqrt{12.25} \ = \ ?\)
\(\displaystyle + \ \sqrt{21} \ < \ \ \sqrt{21.16} \ = \ ?\)
- - - - - - - - - - - - - - - - -

_________________________________________________________________

Or, if you were to do it this way without estimating at all, but with multiple squaring and much
recombining needed (just two beginning steps are shown):

Spoiler

\(\displaystyle \sqrt{6} + \sqrt{12} \ \ \ vs. \ \ \ 11 - \sqrt{21}\)

\(\displaystyle (\sqrt{6} + \sqrt{12})^2 \ \ \ vs. \ \ \ (11 - \sqrt{21})^2\)


Continue on with about 10 steps, plus or minus a few steps.

In your first spoiler 5 is good enough for the upper bound of [imath]\sqrt{21}[/imath].

 

[khansaheb]

Hello!
I need to proof (without computing the square roots!) that "sqrt(6)+sqrt(12)+sqrt(21) < 11". How to do it step by step?

I would take the most direct route:

(√6 + √12 + √21)²

= 6+ 12 + 21 + 2√72 + 2√126 + 2√252

< 39 + 2 * 9 +2*12 + 2*16 = 113 < 121

 

[blamocur]

I would take the most direct route:

(√6 + √12 + √21)²

= 6+ 12 + 21 + 2√72 + 2√126 + 2√252

< 39 + 2 * 9 +2*12 + 2*16 = 113 < 121

It is good, but is it more direct than
[math]\sqrt{6} + \sqrt{12}+\sqrt{21} < \sqrt{6.25}+\sqrt{12.25}+\sqrt{25} = 2.5+3.5+5 = 11[/math] ?

 

[BobP]

Here's another approach,[math]\sqrt{6}+\sqrt{12}+\sqrt{21} \\ = \sqrt{3}(\sqrt{2}+\sqrt{4}+\sqrt{7}) \\ \lt \sqrt{3}(\sqrt{2} +\sqrt{4} +\sqrt{8})[/math]Now if you square, all that you need to say is that [imath]\sqrt{2} \lt1.5[/imath]. That gets you 120.