Comparing $7$ and $\sqrt{2}+\sqrt{5}+\sqrt{11}$: Which is Bigger?

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Discussion Overview

The discussion revolves around comparing the numerical value of $7$ with the expression $\sqrt{2} + \sqrt{5} + \sqrt{11}$. Participants are exploring methods to determine which is larger without the use of a calculator.

Discussion Character

  • Exploratory, Mathematical reasoning

Main Points Raised

  • Participants are tasked with comparing $7$ and $\sqrt{2} + \sqrt{5} + \sqrt{11}$.
  • Some participants suggest proving the comparison without using a calculator.
  • There is mention of using tricks or techniques learned from the forum to approach the problem.
  • Multiple participants express enthusiasm for the challenge and share their intent to use alternative methods.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on which value is larger, and participants have not settled on a definitive method or solution.

Contextual Notes

Participants have not fully detailed their proposed methods or the assumptions underlying their approaches, leaving some steps and reasoning unresolved.

Albert1
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compare :
$7$, and $\sqrt 2+\sqrt 5 + \sqrt {11}$
which one is bigger?
 
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Approximating the roots to 3 significant digits, we have$$\sqrt2+\sqrt5+\sqrt{11}\approx1.41+2.23+3.31=6.95,6.95+3\cdot0.009=6.977\lt7$$
 
greg1313 said:
Approximating the roots to 3 significant digits, we have$$\sqrt2+\sqrt5+\sqrt{11}\approx1.41+2.23+3.31=6.95,6.95+3\cdot0.009=6.977\lt7$$
now ,try to prove it,without using calculator
 
Albert said:
now ,try to prove it,without using calculator

It is fairly easy to compute square roots by hand to two decimal places...[cs=Spiteful]spiteful.gif[/cs]
 
MarkFL said:
It is fairly easy to compute square roots by hand to two decimal places...[cs=Spiteful]spiteful.gif[/cs]
you don't have to compute the square roots of those numbers ,use some tricks
 
I am going to use a trick someone taught me from this forum to solve for this challenge:

Note that

$288<289\,\,\implies2(12^2)<17^2$ or $(\sqrt{2}<\dfrac{17}{12})---(1)$

$80<81\,\,\implies5(4^2)<9^2$ or $(\sqrt{5}<\dfrac{9}{4})---(2)$

$99<100\,\,\implies11(3^2)<10^2$ or $(\sqrt{11}<\dfrac{10}{3})---(3)$

Adding the inequalities in (2), (2) and (3) up gives us the answer:

$\sqrt{2}+\sqrt{5}+\sqrt{11}<\dfrac{17}{12}+\dfrac{9}{4}+\dfrac{10}{3}=7$
 
anemone said:
I am going to use a trick someone taught me from this forum to solve for this challenge:

Note that

$288<289\,\,\implies2(12^2)<17^2$ or $(\sqrt{2}<\dfrac{17}{12})---(1)$

$80<81\,\,\implies5(4^2)<9^2$ or $(\sqrt{5}<\dfrac{9}{4})---(2)$

$99<100\,\,\implies11(3^2)<10^2$ or $(\sqrt{11}<\dfrac{10}{3})---(3)$

Adding the inequalities in (1), (2) and (3) up gives us the answer:

$\sqrt{2}+\sqrt{5}+\sqrt{11}<\dfrac{17}{12}+\dfrac{9}{4}+\dfrac{10}{3}=7$
very smart!
 
Another way.

We have the following equivalences
$$\begin{aligned}\sqrt{2}+\sqrt{5}+\sqrt{11}<7&\Leftrightarrow \sqrt{5}+\sqrt{11}<7-\sqrt{2}\\
&\Leftrightarrow\left(\sqrt{5}+\sqrt{11}\right)^2<\left(7-\sqrt{2}\right)^2 \\&\Leftrightarrow 16+2\sqrt{55}<51-14\sqrt{2}\\
&\Leftrightarrow 2\sqrt{55}+14\sqrt{2}<35\\&\Leftrightarrow \left(2\sqrt{55}+14\sqrt{2}\right)^2<35^2\\&\Leftrightarrow 612+56\sqrt{110}<1225\\
&\Leftrightarrow 56\sqrt{110}<613\\
&\Leftrightarrow \left(56\sqrt{110}\right)^2<613^2\\
&\Leftrightarrow 344960<375769.\end{aligned}$$
 

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