Albert1
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compare :
$7$, and $\sqrt 2+\sqrt 5 + \sqrt {11}$
which one is bigger?
$7$, and $\sqrt 2+\sqrt 5 + \sqrt {11}$
which one is bigger?
now ,try to prove it,without using calculatorgreg1313 said:Approximating the roots to 3 significant digits, we have$$\sqrt2+\sqrt5+\sqrt{11}\approx1.41+2.23+3.31=6.95,6.95+3\cdot0.009=6.977\lt7$$
Albert said:now ,try to prove it,without using calculator
you don't have to compute the square roots of those numbers ,use some tricksMarkFL said:It is fairly easy to compute square roots by hand to two decimal places...[cs=Spiteful]spiteful.gif[/cs]
very smart!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:
$\sqrt{2}+\sqrt{5}+\sqrt{11}<\dfrac{17}{12}+\dfrac{9}{4}+\dfrac{10}{3}=7$