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

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The discussion centers on comparing the value of 7 with the sum of square roots, specifically $\sqrt{2} + \sqrt{5} + \sqrt{11}$. Participants are encouraged to prove which is larger without using a calculator. Some suggest using estimation techniques for the square roots to facilitate the comparison. Others mention employing mathematical tricks learned from the forum to solve the challenge. The conversation emphasizes the importance of analytical thinking in determining the larger value.
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$7$, 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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