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Which side is larger? $$ \sqrt{15} - \sqrt{7} + \sqrt{5} + \sqrt{2} \stackrel{?}{\lessgtr} 5 $$

Without using a calculator, computer, or estimating square roots, please determine which side has the larger value.

Rules of the game: In this puzzle, you have to manipulate inequalities between two sums of radicals (one on the left hand side, and another one on the right hand side). You start with the two expressions $L:=\sqrt{15}-\sqrt{7}+\sqrt{5}+\sqrt{2}$ and $R:=5$ as given above. You are only allowed to perform the following three operations:

  • Add the same value $\Delta$ to $L$ and to $R$, which yields a new left hand side $L+\Delta$ and a new right hand side $R+\Delta$.
  • Multiply both $L$ and $R$ by the same non-negative real number $c$, which yields a new left hand side $c\cdot L$ and a new right hand side $c\cdot R$.
  • Square $L$ and $R$ (given that $L$ and $R$ are non-negative), which yields a new left hand side $L^2$ and a new right hand side $R^2$.

The goal is to reach an inequality with integers on both sides.

There is a "nifty way" of doing this that moves a quantity (a radical) over first. After squaring both sides, and combining integers on respective sides, the integers can be subtracted away. Then you'll be left with the combination of three unlike radicals on one side versus a single fourth unlike radical on the opposite side. But, you should keep going until you have one integer on one side, versus one integer on the other side.

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    $\begingroup$ @Olive: You mean you're posting off-topic questions deliberately because other questions were off-topic? $\endgroup$
    – Deusovi ♦
    Commented Mar 18, 2016 at 5:42
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    $\begingroup$ @2012rcampion - When I posted a similar problem in the Mathematics section, intending for it to be a challenge, they stated it looked as if I posted it as if it were a math problem I was posting without showing any work/attempts of my own. But it isn't that. It's not a homework problem. $\endgroup$
    – Olive Stemforn
    Commented Mar 18, 2016 at 6:21
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    $\begingroup$ @hmmn Something like i.sstatic.net/nB9HK.png ? $\endgroup$
    – 2012rcampion
    Commented Mar 18, 2016 at 6:50
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    $\begingroup$ @hmmn Try $\sqrt{2} + \sqrt{3} + \sqrt{5} \lessgtr \sqrt{6} + \sqrt{7}$, or $\sqrt{2} + \sqrt{3} + \sqrt{5} \lessgtr \sqrt{7} + \sqrt{11}$. I think the properties of this problem are shared by pretty much all sums of radicals. $\endgroup$
    – 2012rcampion
    Commented Mar 18, 2016 at 8:52
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    $\begingroup$ This puzzle is far beyond text book math. There is no general theory for this area. I would expect that the solution will heavily depend on the particular choice of the numbers 15, 7, 5, 2. $\endgroup$
    – Gamow
    Commented Mar 18, 2016 at 12:52

4 Answers 4

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The answer is

The right hand side is bigger

I imagine the line of reasoning the author wants is as follows:

$2352 ~>~ 1$
$\Rightarrow~ 28\sqrt{3} ~>~ 1$
$\Rightarrow~ 3+196+28\sqrt{3} ~>~ 200$
$\Rightarrow~ \sqrt{3}+14 ~>~ 10\sqrt{2}$
$\Rightarrow~ 3+14\sqrt{3} ~>~ 10\sqrt{6}$
$\Rightarrow~ 21+7+14\sqrt{3} ~>~ 10+15+10\sqrt{6}$
$\Rightarrow~ \sqrt{21} + \sqrt{7} ~>~ \sqrt{10} +\sqrt{15}$
$\Rightarrow~ \sqrt{21} -\sqrt{15} + \sqrt{7} ~>~ \sqrt{10}$
$\Rightarrow~ \sqrt{105} -\sqrt{75} + \sqrt{35} ~>~ \sqrt{50}$
$\Rightarrow~ -\sqrt{50} ~>~ -(\sqrt{105} -\sqrt{75} + \sqrt{35})$
$\Rightarrow~ - 2(\sqrt{105} -\sqrt{75} + \sqrt{35}) ~<~- 2\sqrt{50}$
$\Rightarrow~ 27 - 2(\sqrt{105}-\sqrt{75} + \sqrt{35}) ~<~ 27- 2\sqrt{50}$
$\Rightarrow~ 15+7+5 -2(\sqrt{105}-\sqrt{75}+\sqrt{35})~<~25+2-2\sqrt{50}$
$\Rightarrow~ (\sqrt{15} - \sqrt{7} + \sqrt{5})^2 ~<~ (5 - \sqrt{2})^2$
$\Rightarrow~ \sqrt{15} - \sqrt{7} + \sqrt{5} ~<~ 5 - \sqrt{2}$
$\Rightarrow~ \sqrt{15} - \sqrt{7} + \sqrt{5} + \sqrt{2} ~<~ 5$

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    $\begingroup$ A great solution, by the way. The central step seems to be that the value 27 cancels out in lines 4 and 5 from the bottom. $\endgroup$
    – Gamow
    Commented Mar 18, 2016 at 14:19
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    $\begingroup$ Ugh.. somehow I had it in my mind that a negative number squared was also negative otherwise I probably also found it. Nice job! Personally I would write the steps in the other direction though because now you have to read from bottom to top to see what steps you did $\endgroup$
    – Ivo
    Commented Mar 18, 2016 at 14:26
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    $\begingroup$ My insight from this is that in order to solve any future equations of the kind $\sqrt{a}\pm\sqrt{b}\pm\sqrt{c}\pm\sqrt{d} \stackrel{?}{\lessgtr} e$ you just have to calculate $\frac{a+b+c+d-e^2}{2}$ and that will be the square root you have to move to the other side to eliminate a square root $\endgroup$
    – Ivo
    Commented Mar 18, 2016 at 14:38
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    $\begingroup$ @hexomino: I have slightly revised your solution (divided once by sqrt(3) and once by sqrt(5)). It is the same argument, just the numbers are a little bit smaller (and easier to parse). $\endgroup$
    – Gamow
    Commented Mar 18, 2016 at 14:40
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    $\begingroup$ Cheers, hexomino! Cheers, @Gamow! 2012rcampion's treasure map shows that 1 < 2352 is indeed the lowest possible residue. Congratulations as well on keeping the signs straight. $\endgroup$
    – humn
    Commented Mar 18, 2016 at 22:06
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Here's another approach, which starts from smaller integer residuals than hexomino's solution.

$\qquad 14 < 15$
$\Rightarrow \sqrt{14} < \sqrt{15} $
$\Rightarrow 4\sqrt{105} < 30\sqrt2 $
$\Rightarrow 28 + 4\sqrt{105} + 15 < 25 + 30\sqrt2 + 18 $
$\Rightarrow 2\sqrt7 + \sqrt{15} < 5 + 3\sqrt2 \qquad\qquad \text{___ [1]} $

and

$\qquad 35 < 36 $
$\Rightarrow \sqrt{35} < 6 $
$\Rightarrow 6\sqrt{35} < 36 $
$\Rightarrow 32 < 63 - 6\sqrt{35} + 5 $
$\Rightarrow 4\sqrt2 < 3\sqrt7 - \sqrt5 $
$\Rightarrow 3\sqrt2 - \sqrt7 + \sqrt5 < 2\sqrt7 - \sqrt2 \quad \text{___ [2]} $

Finally, bringing these lines of argument togegther,

$2\sqrt7 + \sqrt{15} + 3\sqrt2 - \sqrt7 + \sqrt5 < 2\sqrt7 + \sqrt{15} + 2\sqrt7 - \sqrt2 \quad \text{by [2]} $ $ \qquad\qquad < 5 + 3\sqrt2 + 2\sqrt7 - \sqrt2 \quad \text{by [1]} $
$\Rightarrow \sqrt{15} - \sqrt7 + \sqrt5 < 5 - \sqrt2 $
$\Rightarrow \sqrt{15} - \sqrt7 + \sqrt5 + \sqrt2 < 5 $

I admit it isn't a single chain of inferences.

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Yet another method:

$$105 > 49 \Rightarrow \sqrt{105} > 7 \Rightarrow 22 +2\sqrt{105} > 22+14$$ $$\Rightarrow (\sqrt{15})^2 + (\sqrt{7})^2 + 2 \sqrt{15} \sqrt{7} > 36$$ $$\Rightarrow \sqrt{15} + \sqrt{7} > 6 \tag{1}$$

And then:

$$9^2 \times 10 < 29^2 \Rightarrow \sqrt{10} < \frac{29}{9} \Rightarrow 7 + 2 \sqrt{10} < \frac{29 \cdot 2 + 7 \cdot 9}{9}$$ $$\Rightarrow (\sqrt{5})^2 + (\sqrt{2})^2 + 2\sqrt{10} < \frac{121}{9}$$ $$\Rightarrow \sqrt{5} + \sqrt{2} < \frac{11}{3} \Rightarrow \frac{8}{6} +\sqrt{5} + \sqrt{2} < 5$$ $$\Rightarrow \frac{15-7}{\sqrt{15} + \sqrt{7}} +\sqrt{5} + \sqrt{2} < 5 \tag{*}$$ $$\Rightarrow \sqrt{15} - \sqrt{7} +\sqrt{5} + \sqrt{2} < 5$$

(*): From (1), increasing the denominator decreases the fraction; strictly smaller than LHS of previous step

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  • $\begingroup$ No problem! The advantage of this method is that the numbers are a lot smaller. $\endgroup$
    – Toby Mak
    Commented Jul 16, 2021 at 0:28
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The right side is larger because

You square the roots and do the sum, which equals 15 and you suqare the right side which gives you 25.

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    $\begingroup$ $a^2 + b^2 \neq (a+b)^2$. $\endgroup$
    – Deusovi ♦
    Commented Mar 17, 2016 at 23:28
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    $\begingroup$ Not exactly the right method... $\endgroup$
    – Sid
    Commented Jul 16, 2016 at 12:08

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