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prove 7 > (2^0.5 + 5^0.5 + 11^0.5)

Albert

Well-known member
Jan 25, 2013
1,225
originated from checkittwice

please prove :

$7>(\sqrt 2+\sqrt 5 +\sqrt {11}\, )$

note : you can not use calculator or computer
 
Last edited:

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Re: prove 7>(2^0.5+5^0.5+11^0.5)

$$7 > \sqrt{2} + \sqrt{5} + \sqrt{11}$$

$$7 - \sqrt{11} > \sqrt{2} + \sqrt{5}$$

$$ \left (7 - \sqrt{11} \right )^2 > \left ( \sqrt{2} + \sqrt{5} \right )^2$$

$$60 - 14 \sqrt{11} > 7 + 2 \sqrt{10}$$

$$53 > 2 \sqrt{10} + 14 \sqrt{11}$$

$$53 - 14 \sqrt{11} > 2 \sqrt{10}$$

$$\left ( 53 - 14 \sqrt{11} \right )^2 > \left ( 2 \sqrt{10} \right )^2$$

$$4965 - 1484 \sqrt{11} > 4 \cdot 10$$

$$4925 > 1484 \sqrt{11}$$

$$4925^2 > 1484^2 \cdot 11$$

$$24255625 > 24224816$$

Very ugly, but at least it does not calculate any square root. I remember proving this more elegantly before... (Worried)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: prove 7>(2^0.5+5^0.5+11^0.5)

A very clever method posted at MMF, which I wanted to share here, is as follows:

$\displaystyle 7=\frac{17}{12}+\frac{9}{4}+\frac{10}{3}=\sqrt{ \frac{289}{144}}+\sqrt{\frac{81}{16}}+\sqrt{\frac{100}{9}}=\sqrt{2+\frac{1}{144}}+\sqrt{5+\frac{1}{16}}+\sqrt{11+\frac{1}{9}}>\sqrt{2}+\sqrt{5}+\sqrt{11}$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Re: prove 7>(2^0.5+5^0.5+11^0.5)

$$7 > \sqrt{2} + \sqrt{5} + \sqrt{11}$$

$$7 - \sqrt{11} > \sqrt{2} + \sqrt{5}$$

$$ \left (7 - \sqrt{11} \right )^2 > \left ( \sqrt{2} + \sqrt{5} \right )^2$$

$$60 - 14 \sqrt{11} > 7 + 2 \sqrt{10}$$

$$53 > 2 \sqrt{10} + 14 \sqrt{11}$$

$$53 - 14 \sqrt{11} > 2 \sqrt{10}$$

$$\left ( 53 - 14 \sqrt{11} \right )^2 > \left ( 2 \sqrt{10} \right )^2$$

$$4965 - 1484 \sqrt{11} > 4 \cdot 10$$

$$4925 > 1484 \sqrt{11}$$

$$4925^2 > 1484^2 \cdot 11$$

$$24255625 > 24224816$$

Very ugly, but at least it does not calculate any square root. I remember proving this more elegantly before... (Worried)
Actually, your logic has to start at the bottom and go to the top. In that direction, you are taking square roots.
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Re: prove 7>(2^0.5+5^0.5+11^0.5)

Actually, your logic has to start at the bottom and go to the top. In that direction, you are taking square roots.
How do you mean? I never actually calculated a square root, just manipulating them as $\sqrt{a} \sqrt{b} = \sqrt{ab}$ and similar rules.

Besides, wouldn't going backwards require you to solve $a^2 = 24255625$?

EDIT: ah, I see what you mean. But why isn't it valid top-to-bottom? I thought every step was an equivalence?
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Re: prove 7>(2^0.5+5^0.5+11^0.5)

@Bacterius:

In thinking about your logic going from bottom to top, I discovered that it's not the square roots that are the "problem". The rule for inequalities is this: any time you take a monotonically increasing function and apply it to both sides of an inequality, you get to preserve the direction of the inequality. Any time you take a monotonically decreasing function and apply it to both sides of an inequality, you need to reverse the direction of the inequality. If the function is not monotonically increasing or decreasing, then watch out!

The square root function is monotonically increasing on its domain. So, no problem. The square function, on the other hand, is monotonically decreasing for negative inputs, and monotonically increasing for positive outputs. So any time you square something in an inequality proof, you need to check that both sides are positive (keep the direction of the inequality) or that they are both negative (reverse the inequality).

Now, if I start at the bottom of your proof and work my way up, I notice that there are only three different kinds of steps in your proof: rewrites, square roots, and "shuffles" (shifting things from one side to the other by using addition or subtraction, which never requires an inequality reversal). You never square anything. So I think your proof is fine.

I think in my mind I was wondering about square roots, because if you start from $x^{2}>y^{2}$, you could either have $x>y$ or $-x<-y$. But in your proof, there are no variables anywhere, it's all numbers. And if you start from the bottom and work your way up, you can see that you take the positive square root a number of times, which is perfectly fine.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Re: prove 7>(2^0.5+5^0.5+11^0.5)

Sorry to invade this thread (Tmi), but I'd like to answer Bacterius' question:

Bacterius said:
But why isn't it valid top-to-bottom? I thought every step was an equivalence?

If you go top-to-bottom you're essentially assuming what you want to prove, arriving at some true statement and saying "therefore, what we assumed is right", and you can't do that. They may be equivalent, but the order which you show they are equivalent is important.

Again, sorry for the intrusion.
(Tmi)

Cheers.