I've found this inequality, it seems cute:

$$\sqrt{2} + 2 \sqrt{3} + 2 \sqrt{5} + \sqrt{7} < 12$$

Note: It checks with a calculator. I am also interested in some methods that work for more general sum of radicals.

Thank you for your interest!

$\bf{Added:}$ I found this inequality by some random search. Note that all of the coefficients on LHS are positive, this can be an advantage. In the method of @heropup, one wants to check that a certain integer $N$ ( here $N=12$) is larger than a totally real algebraic number $\alpha$ that is larger than all the other conjugates $\alpha_j$. Say its minimal polynomial is $f(x)$. Then we need to show that $N>$ all of the roots of $f(x)$. Note that the roots of all derivatives of $f$ are also between the largest and the smallest of $\alpha_j$'s. The above inequality is equivalent to $f(N)$, $f'(N)$, $f''(N)$, $\ldots > 0$. With Taylor's formula, this is equivalent to: the polynomial $f(N+t)$ has all of the coefficients positive in $t$. This is an automated check that can be done instantly after we found $f(x)$.

This method does not work if $\alpha$ is not the largest ( or smallest) of all its conjugates. But there is a way to "make it the larges" using a transformation $x \mapsto 1/(a-x)$. I will leave out the details.

There was a previous question on this site about finding a general method for comparing arbitrary real algebraic numbers. That seems like a quite involved problem. It is easy to compare with $0$ a real algebraic number $\alpha$ with all its other conjugates non-real. The test is simple $\alpha> 0$ if and only if $N(\alpha) > 0$, where $N(\cdot)$ is the norm. In particular this applies to cubic fields with discriminant $<0$. The general problem of comparing with $0$ an algebraic number $\alpha$ is simple if $\alpha$ is the $k$-th smallest ( or largest) root of an irreducible polynomial. Indeed, one needs to see how many positive/negative roots does the polynomial have , something that can be decided ( in principle). This is harder if one is given an expression involving several real algebraics. Even if we can find an equation whose root is the given expression, it could be harder to decide the order rank of the roots. So this seems like a difficult problem. But perhaps it is solved somewhere in the literature.

One other approach of @Parcly Taxel: simply gets rid of all the roots, little by little, something that I thought not possible just by counting. That is interesting! I am not fully convinced that it is so for an arbitrary number of roots as I was assured by Parcly... tempting to through another challenge to see how that works, what is the logic there...

Another approach, of @Luke Collins, and @Claude Leibovici uses the approximation of square roots with their convergents. It turns out that the convergents are easy to form, since the continued fractions are periodic ( and in this case, quite simple).

Summing up: some methods are ingenious and algebraic, other use some the minimal polynomial, other use continued fractions and convergents. Perhaps I should mention that there are other way to approximate roots. Parcly seems to prefer power series. I like Newton's method. Claude Leibovici might like Pade approximants too.

I've learned a lot from All the answers! Thank you very much to All the contributors!

  • 3
    $\begingroup$ The sum is very close to $12$ , so to prove this without a calculator won't be quite easy. $\endgroup$
    – Peter
    Feb 18 at 8:27
  • $\begingroup$ @Riemann: In case of $4$ or more radicals, raising to the square does not seem to eliminate the radicals. $\endgroup$
    – orangeskid
    Feb 18 at 8:41
  • $\begingroup$ @Riemann: It would be great if you would sketch your approach below in an answer ? Thank you for your feedback! $\endgroup$
    – orangeskid
    Feb 18 at 8:49
  • 1
    $\begingroup$ I see now that the squaring do not eliminates radicals easily, so I am trying to find other method. If I find one, I will post it as an answer. $\endgroup$
    – Riemann
    Feb 18 at 9:01
  • 1
    $\begingroup$ My idea was to use the fact that $\sqrt x\leqslant \min\big(\frac{1+x}2, \frac{4+x}4, \frac{9+x}6\big)$, but this only allows you to get $<73/6\approx 12.1$. You'd think it would be enough! i.stack.imgur.com/44Flx.png $\endgroup$ Feb 18 at 9:34

8 Answers 8


Rewrite the LHS as $$\frac43\sqrt{1+\frac18}+4\sqrt{1-\frac14}+4\sqrt{1+\frac14}+\frac83\sqrt{1-\frac1{64}}$$ and use the binomial series to derive upper bounds for each term. For the series corresponding to $2\sqrt3$ and $\sqrt7$ every partial sum is an upper bound since all terms after the zeroth are negative. For the other two series only the partial sums up to odd-indexed terms (even number of terms total) form valid upper bounds since those series alternate and odd-indexed terms are positive.

Taking four terms (up to index $3$) for the first three series and just two (up to index $1$) for the $\sqrt7$ series yields an upper bound of $$\frac{8689}{2^{11}×3}+\frac{887}{2^8}+\frac{1145}{2^8}+\frac{127}{2^4×3}<12$$ where the denominators have been factorised to emphasise the relative ease of doing this by hand.

  • 1
    $\begingroup$ Very interesting. Do roots always disappear no matter how many you start with? $\endgroup$
    – orangeskid
    Feb 18 at 9:23
  • $\begingroup$ [+1] You forgot two square roots on the first line $\endgroup$ Feb 18 at 9:28
  • $\begingroup$ The approximation method is very interesting! I wonder if other approximations of roots are not preferred, like Newton's method, or maybe Pade? $\endgroup$
    – orangeskid
    Feb 18 at 10:09
  • $\begingroup$ @orangeskid Yes, repeated squaring will eventually get you the answer in rational arithmetic as long as all roots are of integers. The binomial series is the simplest approach to inequality problems with a bunch of roots – no reason to use the harder Newton's method. $\endgroup$ Feb 18 at 10:11
  • 2
    $\begingroup$ @ParclyTaxel Repeated squaring will not work when we have 5+ radicals (involving 5+ primes). EG Can you show that $ \sqrt{2} + \sqrt{3} + \sqrt{5} + \sqrt{7} + \sqrt{11} + \sqrt{13} < 15$. Unfortunately you can't get things to cancel out (and they might get much uglier) $\endgroup$
    – Calvin Lin
    Feb 18 at 13:38

While extremely tedious to do by hand calculation, a general approach would be to compute the minimal polynomial of $\mathbb Q[\sqrt{2} + 2\sqrt{3} + 2\sqrt{5} + \sqrt{7}]$, which is

$$f(x) = x^{16}-328 x^{14}+38396 x^{12}-2060088 x^{10}+54775910 x^8-726131832 x^6+4479958076 x^4-10024901512 x^2+489692641.$$

Then evaluating $f(12) = 22250268794209 > 0$ proves the inequality.

  • $\begingroup$ Yes, that was my approach too. Well, I did check that $f > 0$ on $[12, +\infty)$ by the subst $x=12 + t$, $t\ge 0$, or other arguments needed. $\endgroup$
    – orangeskid
    Feb 18 at 9:20
  • 1
    $\begingroup$ This is a nice one $\endgroup$ Feb 18 at 10:07

As you suggested (in response to my original answer below), you can take better and better convergents of the individual square roots and get an upper-bound that way. This will certainly be easier by hand, since they are periodic. For example,

\begin{align*} \sqrt 2 &= 1 + \frac{1}{1+\sqrt 2} \\[3pt] &= 1 + \frac1{2+\frac{1}{1+\sqrt 2}}\\[3pt] &= 1 + \frac{1}{2+ \frac{1}{2+\cdots}} \\[3pt] & < 1 + \frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}} &&\text{( $<$ since odd convergent)}\\ &= \frac{99}{70}. \end{align*}

@Claude Leibovici seems to have done this in his answer.

Indeed, using this idea, you get \begin{align*} \sqrt{2}+2 \sqrt{3}+2 \sqrt{5}+\sqrt{7} &<\frac{99}{70}+2\left(\frac{97}{56}\right)+2\left(\frac{161}{72}\right)+\frac{127}{48}\\[3pt]&=\frac{60\,463}{5040} <\frac{60\,480}{5040} = 12. \end{align*}

Original answer: I don't know if you'd like this, but you could argue that the convergents of a continued fraction alternate above/below their limit, and the first few convergents of your number are $$11, 12, \frac{3155}{263},\frac{9477}{790}$$ so $\sqrt2+2\sqrt3+2\sqrt5+\sqrt7<\frac{9477}{790}<12$.

(Recall the first few convergents of $n$ are $$\lfloor n\rfloor, \lfloor n\rfloor +\frac{1}{\left\lfloor \frac{1}{n-\lfloor n\rfloor }\right\rfloor }, \lfloor n\rfloor +\frac{1}{\left\lfloor \frac{1}{n-\lfloor n\rfloor }\right\rfloor +\frac{1}{\left\lfloor \frac{1}{\frac{1}{n-\lfloor n\rfloor }-\left\lfloor \frac{1}{n-\lfloor n\rfloor }\right\rfloor }\right\rfloor }}, \text{etc.},$$ a calculator is handy to help with computing these).

  • $\begingroup$ What if we use the convergents of each square roots, since they are periodic and easy to get, then take some convergents that are larger for each of them. Otherwise do we have a good algorithm for the continued fraction of this number? $\endgroup$
    – orangeskid
    Feb 18 at 9:47
  • $\begingroup$ @orangeskid That will probably work actually, if you take enough of them for each square root! $\endgroup$ Feb 18 at 9:47
  • 1
    $\begingroup$ The continued fraction of $\sqrt{d}- [\sqrt{d}]$ is simply periodic, quite simple for the above, just checked on WA, not sure how many reduced are necessary. $\endgroup$
    – orangeskid
    Feb 18 at 9:51

Here's a solution that can actually be done by hand without too much work.

First off we have the inequalities $$k-\frac{1}{2k}-\frac{1}{4k^3}<\sqrt{k^2-1}<k-\frac{1}{2k}$$ for $k>2$.
Next note the remarkable identity $$ \sqrt{24}(\sqrt{2} + 2\sqrt{3} + 2\sqrt{5} + \sqrt{7}) = \sqrt{7^2-1}+\sqrt{17^2-1} + 2\sqrt{11^2-1} + \sqrt{13^2-1} $$ Thus multiplying both sides by $\sqrt{24}$ gives the inequality $$ \sqrt{7^2-1}+\sqrt{17^2-1} + 2\sqrt{11^2-1} + \sqrt{13^2-1} < 12 \sqrt{5^2-1} $$ For the LHS we have \begin{eqnarray*} \sqrt{7^2-1}+\sqrt{17^2-1} + 2\sqrt{11^2-1} + \sqrt{13^2-1} &<& 7-\frac{1}{2*7}+13-\frac{1}{2*13}\\ & & +17-\frac{1}{2*17}+2*11-\frac{2}{2*11}\\ &=& 59 -\frac{7835}{34034}\\ &=& 58+\frac{26199}{34034}. \end{eqnarray*} This calculation is not too hard when you note that $(7*11*13)=1001$, so the denominator readily becomes $(2*17)*1001=34034$.

For the RHS we have the following bound \begin{eqnarray*} 12\sqrt{5^2-1} &>& 12 (5-\frac{1}{2*5}-\frac{1}{4*5^3})\\ &=& 60- \frac{153}{125}\\ &=& 58+\frac{97}{125} \end{eqnarray*} We can then readily verify that $\frac{26199}{34034}<\frac{97}{125}$ by cross multiplying by $125*34034$ giving $$ 26199*125 = 3274875 < 97 * 34034 = 3301298 $$ which is clearly true. To verify that my hand calculations are correct we can evaluate all the expressions numerically with a calculator. We have \begin{eqnarray*} \mathrm{LHS}=58.769149675775\ldots &<&58+\frac{26199}{34034} \\ & = & 58.769{\color{red}{7890344949\ldots}}\\ &< & 58+\frac{97}{125}\\ &=& 58.7{\color{red}{76}}\\ &<& 12\sqrt{24} \\ &=& 58.7{\color{red}{877538267963\ldots}}=\mathrm{RHS} \end{eqnarray*}

  • $\begingroup$ Lovely idea! Quite an alignment... Using a calculator I noticed that already $26200/34000< 97/125$, that could simplify a bit at the end. Now, the true error is about $3.7/1000$, and after multiplying by $\simeq 5$ it becomes $1.8/100$, while your estimates are around $1.2/100$, so very plausible. Great! $\endgroup$
    – orangeskid
    Feb 22 at 7:25

We'll use the Newton recurrence

$$x \mapsto f_a(x) = \frac{1}{2}(x + \frac{a}{x})$$ to approximate $\sqrt{a}$. Note that for all $x> 0$, we have $f_a(x)\ge \sqrt{a}$, and if $x> \sqrt{a}$, then $x > f_a(x)> \sqrt{a}$. Moreover, the convergence of the sequence $(f^n_a(x))_n$ towards $\sqrt{a}$ is fast.

So we'll approximate LHS $\sqrt{2} + 2 \sqrt{3} + 2 \sqrt{5} + \sqrt{7}$ from above with

$$f_2^3(2) + 2 f_3^2(2) + 2 f_5^2(2) + f_7^2(2)= \frac{577}{408} + 2 \cdot \frac{97}{56} + 2 \cdot \frac{161}{72} + \frac{233}{88}= \frac{565415}{47124} < 12 $$

One advantage of this method is that the fact that upper bound is automatic. In fact this upper bound will work for radicals of any order $n$ ( by the Newton's tangent method the iteration function for the equation $x^n - a= 0 $ is $f_{a, n}(x) = \frac{1}{n} ( (n-1) x + \frac{a}{x^{n-1}})$

Note that any inequality with radicals can be written in the form

$$\sum \frac{c_i}{\sqrt{a_i}} > \sum c'_j \sqrt{a'_j}$$

or even for radicals of higher orders, and the same idea could work in principle.


Alternative solution:

We have \begin{align*} (\sqrt 3 + \sqrt 5)^2 &= 8 + 2\sqrt{15}\\ &= 8 + 2\sqrt{16} - 2(\sqrt{16} - \sqrt{15})\\ &= 16 - 2\cdot \frac{16 - 15}{\sqrt{16} + \sqrt{15}}\\ &< 16 - 2\cdot \frac{16 - 15}{\sqrt{16} + \sqrt{16}}\\ &= \frac{63}{4} \end{align*} which results in $\sqrt 3 + \sqrt 5 < \frac32 \sqrt 7$ and $2\sqrt 3 + 2\sqrt 5 < 3\sqrt 7$.

We have $$\sqrt 2 + 2\sqrt 3 + 2\sqrt 5 + \sqrt 7 < \sqrt 2 + 4\sqrt 7 < 12$$ where we use $12^2 - (\sqrt 2 + 4\sqrt 7)^2 = 144 - (114 + 8\sqrt{14}) = 2(15 - 4\sqrt{14}) > 0$ (using $15^2 > 16\times 14$).

We are done.


Using rational numbers $$\sqrt 2=\frac{99}{70}\qquad \sqrt 3=\frac{97}{56}\qquad \sqrt 5=\frac{161}{72}\qquad \sqrt 7=\frac{127}{48}$$ $$\frac{99}{70}+2\times\frac{97}{56}+2\times\frac{161}{72}+\frac{127}{48}=\frac{60463}{5040}$$

I used my phone for the rhs. $$60463 < 12 \times 5040=60480$$

Pure luck !

  • $\begingroup$ This is similar to what OP suggested in response to my answer. Nice! $\endgroup$ Feb 18 at 9:51
  • $\begingroup$ @LukeCollins. I did not see your answer (since typing). $\endgroup$ Feb 18 at 9:54
  • $\begingroup$ Those are convergents... (so I want the app he's having :-) ) So luck comes to the ones prepared. $\endgroup$
    – orangeskid
    Feb 18 at 9:59
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    $\begingroup$ The equalities in the first line are of course all wrong. Why don't you write $<$ instead, to make this a correct proof? $\endgroup$
    – Martin R
    Feb 18 at 10:42

Hint : Using Cauchy Schwarz inequality we have :


Hint for the end after MartinR comment we have :


Bonus inequality :

For $0<x\leq 1$ we have :

$$x^{\frac{2}{9}}\left(\sqrt{2}+\sqrt{12}+\sqrt{20}+\sqrt{7}\right)< 6+\frac{1}{2}\left(\sqrt{1+e^{\frac{2}{4}x}}+2\sqrt{1+e^{\frac{3}{4}x}}+2\sqrt{1+e^{\frac{5}{4}x}}+\sqrt{1+e^{\frac{7}{4}x}}\right)<12$$

  • $\begingroup$ How do you prove (without calculator) that $\left(\left(\sqrt{2}+\sqrt{20}\right)^{2}+\left(\sqrt{12}+\sqrt{7}\right)^{2}\right)2<144$? $\endgroup$
    – Martin R
    Feb 18 at 16:08
  • $\begingroup$ @MartinR see my edit for you and someone else . $\endgroup$ Feb 18 at 16:51
  • $\begingroup$ Interesting... a bit hard for me to follow the arguments... maybe there is a general fact there... $\endgroup$
    – orangeskid
    Feb 18 at 23:25

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