Show that $\sqrt{10}+\sqrt{26}+\sqrt{17}+\sqrt{37} \gt \sqrt{341}$

Question

Show that $\sqrt{10}+\sqrt{26}+\sqrt{17}+\sqrt{37} \gt \sqrt{341}$.

This is inspired by Showing $x+y>z$, where $x=\sqrt{10}+\sqrt{26}$, $y=\sqrt{17}+\sqrt{37}$, and $z=\sqrt{323}$. Is my idea corect?, where the 341 is replaced by 323.

In that problem, the difference is about $0.4949$, which enabled a quite elementary proof to work.

In this case, the difference is about $0.00098$, which is much harder.

So, is an elementary proof possible, other than computing the difference?

Note: Wolfram Alpha gives this form for the difference:

sqrt(root of x^16 - 6896 x^15 + 21218584 x^14 - 38619086608 x^13 + 46445175324092 x^12 - 39034285182032752 x^11 + 23634682317529311848 x^10 - 10471213870456147495696 x^9 + 3411556529576995933189478 x^8 - 814131450981226210018475344 x^7 + 140459189711872042665929874728 x^6 - 17103305259239135613970718210992 x^5 + 1412793771745512798455228682417916 x^4 - 74118197304168530774085170831631440 x^3 + 2187202048899771587108104647206992600 x^2 - 27077232770375735729098901781263934000 x + 26005877616308367788704404950625 near x = 9.60433×10^-7)

It also gives the minimal polynomial as

x^32 - 6896 x^30 + 21218584 x^28 - 38619086608 x^26 + 46445175324092 x^24 - 39034285182032752 x^22 + 23634682317529311848 x^20 - 10471213870456147495696 x^18 + 3411556529576995933189478 x^16 - 814131450981226210018475344 x^14 + 140459189711872042665929874728 x^12 - 17103305259239135613970718210992 x^10 + 1412793771745512798455228682417916 x^8 - 74118197304168530774085170831631440 x^6 + 2187202048899771587108104647206992600 x^4 - 27077232770375735729098901781263934000 x^2 + 26005877616308367788704404950625

Answer 1

Let $$ s \doteq \sqrt{10} + \sqrt{17} + \sqrt{26} + \sqrt{37} = \sum_{n=3}^6 \sqrt{n^2+1}. $$ The inequality $s > \sqrt{341}$ is fairly tight because $s^2 \approx 341.0362$. However, we will obtain the bound $s^2 > 341.03$ using only rational arithmetic.

To begin, note that $$ \sqrt{n^2+1} - n = \frac{1}{\sqrt{n^2+1} + n} < \frac{1}{2n}. $$ The inequality goes the wrong way to use directly. Instead we express it as $\sqrt{n^2+1} < n + \frac{1}{2n}$ and write $$ \sqrt{n^2+1} - n = \frac{1}{\sqrt{n^2+1} + n} > \frac{1}{n + \frac{1}{2n} + n} = \frac{2n}{4n^2+1}. $$ Therefore $$ s = \sum_{n=3}^6 \sqrt{n^2+1} > \sum_{n=3}^6 \left(n + \frac{2n}{4n^2+1}\right) = \frac{130086126}{7044245}. $$ The final step is just some more rational arithmetic (albeit with large integers!): $$ s^2 - 341 = \frac{1506999259351}{49621387620025} > \frac{15}{500} = 0.03. $$ Therefore $s^2 > 341.03$.

Answer 2

Here is a proof which only contains small numbers.

First of all, we will prove that

$\sqrt{n^2+1}>n+\dfrac1{2n}-\dfrac1{8n^3}\;\;\,$ for any $\,n\in\Bbb N\,.\quad\color{blue}{(1)}$

For any $\,n\in\Bbb N\,$ it results that

$\begin{align}\sqrt{n^2+1}&=\sqrt{\left(n+\dfrac1{2n}\right)^2\!-\dfrac1{4n^2}}=\\[3pt]&=\sqrt{\left(n+\dfrac1{2n}\right)^2\!-\dfrac1{4n^3}\left(n+\dfrac1{2n}\right)+\dfrac1{8n^4}}=\\[3pt]&=\sqrt{\left(n\!+\!\dfrac1{2n}\right)^2\!\!-\!\dfrac1{4n^3}\left(n\!+\!\dfrac1{2n}\right)\!+\!\dfrac1{64n^6}\!+\!\dfrac1{8n^4}\!-\!\dfrac1{64n^6}}=\\[3pt]&=\sqrt{\left(n+\dfrac1{2n}-\dfrac1{8n^3}\right)^2\!+\dfrac1{64n^6}\left(8n^2-1\right)}>\\[3pt]&>\sqrt{\left(n+\dfrac1{2n}-\dfrac1{8n^3}\right)^2}=n+\dfrac1{2n}-\dfrac1{8n^3}\;.\end{align}$

By applying the inequality $\,(1)\,$ we get that

$\sqrt{10}+\sqrt{17}+\sqrt{26}+\sqrt{37}=$

$=\sqrt{3^2+1}+\sqrt{4^2+1}+\sqrt{5^2+1}+\sqrt{6^2+1}=$

$=\displaystyle\!\!\sum\limits_{n=3}^6\sqrt{n^2+1}>\!\sum\limits_{n=3}^6\left(n+\dfrac1{2n}-\dfrac1{8n^3}\right)=$

$=\displaystyle18+\dfrac12\sum\limits_{n=3}^6\dfrac1{n}-\dfrac18\sum\limits_{n=3}^6\dfrac1{n^3}>$

$>18+\dfrac12\!\cdot\!\dfrac{19}{20}-\dfrac18\left(\dfrac1{27}+\color{brown}{\dfrac1{60}}+\color{brown}{\dfrac1{120}}+\dfrac1{216}\right)=$

$\require{cancel}=18+\dfrac{19}{40}-\dfrac18\left(\dfrac{\color{#8888FF}{\cancel{\color{black}9}}^1}{\color{#8888FF}{\cancel{\color{black}{216}}}_{24}}+\color{brown}{\dfrac{\color{#8888FF}{\cancel{\color{brown}3}}^1}{\color{#8888FF}{\cancel{\color{brown}{120}}}_{40}}}\right)=$

$=18+\dfrac{19}{40}-\dfrac1{\color{#8888FF}{\cancel{\color{black}{8}}}_{\!1}}\!\cdot\!\dfrac{\color{#8888FF}{\cancel{\color{black}8}}^1}{120}=18+\dfrac{\color{#8888FF}{\cancel{\color{black}{56}}}^7}{\color{#8888FF}{\cancel{\color{black}{120}}}_{\!15}}=\dfrac{277}{15}>$

$>\sqrt{\dfrac{277^2-2^2}{15^2}}=\sqrt{\dfrac{\big(277+2\big)\big(277-2\big)}{3^2\cdot5^2}}=$

$=\sqrt{\dfrac{\color{#8888FF}{\cancel{\color{black}{279}}}^{31}\!\!\!\!\cdot\!\color{#8888FF}{\cancel{\color{black}{275}}}^{11}}{\color{#8888FF}{\cancel{\color{black}9}}_{\!1}\cdot\color{#8888FF}{\cancel{\color{black}{25}}}_{\!1}}}=\sqrt{341}\;.$

Answer 3

This is actually an answer. We may exploit the fact that for any $n\in\mathbb{R}^+$ the irrational number $\sqrt{n^2+1}$ has a very simple continued fraction, namely $\sqrt{n^2+1}=[n;\overline{2n}]$. By expanding up to the third term we get

$$ \sqrt{10} > \frac{117}{37},\quad \sqrt{17}>\frac{268}{65},\quad \sqrt{26}>\frac{515}{101},\quad \sqrt{37}>\frac{882}{145} $$ and the square of the sum of these fractions exceeds $341$ by approximately $\frac{3}{100}$.

Answer 4

Not an answer, but too long for a comment. The slightly weaker inequality $$\sqrt{10}+\sqrt{17}+\sqrt{26}+\sqrt{37} > \sqrt{53} + 5\sqrt{5} $$ is easily proved by a simple geometric argument. We consider a convex quadrilateral $ABCD$ with vertices at $A(3,1)$, $B(-1,6)$, $C(-4,1)$, $D(1,-5)$ and recall than $\min\left(PA+PB+PC+PD\right)$ is attained at $P=AC\cap BD$. The LHS is $OA+OB+OC+OD$ and the RHS is $PA+PB+PC+PD=AC+BD$.

Answer 5

We may square both sides.

It suffices to prove that $$\sqrt{10\cdot 17} + \sqrt{10\cdot 26} + \sqrt{10 \cdot 37} + \sqrt{17 \cdot 26} + \sqrt{17 \cdot 37} + \sqrt{26 \cdot 37} > 251/2.$$

We have $$\sqrt{10\cdot 17} - 13 = \frac{1}{\sqrt{10\cdot 17} + 13} > \frac{1}{14 + 13} = \frac{1}{27}, $$ $$\sqrt{10\cdot 26} - 16 = \frac{4}{\sqrt{10\cdot 26} + 16} > \frac{4}{17 + 16} = \frac{4}{33},$$ $$\sqrt{10\cdot 37} - 19 = \frac{9}{\sqrt{10\cdot 37} + 19} > \frac{9}{20 + 19} = \frac{9}{39},$$ $$\sqrt{17\cdot 26} - 21 = \frac{1}{\sqrt{17\cdot 26} + 21} > \frac{1}{22 + 21} = \frac{1}{43},$$ $$\sqrt{17\cdot 37} - 25 = \frac{4}{\sqrt{17\cdot 37} + 25} > \frac{4}{26 + 25} = \frac{4}{51},$$ $$\sqrt{26\cdot 37} - 31 = \frac{1}{\sqrt{26\cdot 37} + 31} > \frac{1}{32 + 31} = \frac{1}{63}.$$ (Note: For example, $\sqrt{26\cdot 37} = \sqrt{(5^2 + 1)(6^2 + 1)} \ge 5\cdot 6 + 1 = 31$ by C-S. So we estimate $\sqrt{26\cdot 37} - 31$.)

It suffices to prove that $$\dfrac{1}{27} + \dfrac{4}{33} + \dfrac{9}{39} + \dfrac{1}{43} + \dfrac{4}{51} + \dfrac{1}{63} > \dfrac{1}{2}$$ which is true. (Is there an easy way to prove it not via brute force?)

We are done.

Answer 6

Since some people are comfortable with large polynomials, we can use a method that (barely) works in this situation. We are to show that

$$\sqrt{\frac{10}{341}} + \sqrt{\frac{26}{341}}+\sqrt{\frac{17}{341}}+\sqrt{\frac{37}{341}}-1 > 0$$

Now LHS is the largest root of a polynomial of degree $16$ with all roots real. We could write it down, but the important thing is this: the constant term is $< 0$, and the leading term is positive. It follows that it has a positive root. So the largest root ( the LHS) is $> 0$.

$\bf{Added:}$ A bit more work shows similarly that

$$\sqrt{341 + \frac{37}{1000}} > \sqrt{10} + \sqrt{17} + \sqrt{26} + \sqrt{37} > \sqrt{341 + \frac{36}{1000}}$$

Answer 7

Using the same idea as @Jim Ferry $$s = \sum_{n=3}^6 \sqrt{n^2+1} > \sum_{n=3}^6 \left(n + \frac{4 n \left(2 n^2+1\right)}{16 n^4+12 n^2+1}\right) = \frac{24266790219724938}{1314050635378105}$$ $$\frac{24266790219724938}{1314050635378105} >\frac{277}{15}$$ $$\left(\frac{277}{15}\right)^2=341+\frac{4}{225}$$