Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values

Question

I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.

I tried to apply square on both side but it still carries the root terms.

Any trick I was missing here?

Answer 1

Note that $8>\sqrt{11}$. Therefore\begin{align}\sqrt5+\sqrt6+\sqrt{11}>8&\iff\sqrt5+\sqrt6>8-\sqrt{11}\text{ ($3$ square roots)}\\&\iff11+2\sqrt{30}>75-16\sqrt{11}\text{ (only $2$ square roots)}\\&\iff\sqrt{30}>32-8\sqrt{11}\\&\iff30>1728-512\sqrt{11}\text{ (a single square root)}\\&\iff512\sqrt{11}>1698\\&\iff256\sqrt{11}>849,\end{align}which is true, since $256^2\times11=720\,896$ and $849^2=720\,801$.

Answer 2

We'll prove that: $$\sqrt5+\sqrt6>8-\sqrt{11}$$ or $$11+2\sqrt{30}>64-16\sqrt{11}+11$$ or $$\sqrt{15}+4\sqrt{22}>16\sqrt2$$ or $$15+352+8\sqrt{15\cdot22}>512$$ or $$8\sqrt{15\cdot22}>145$$ or $$8\sqrt{66}>29\sqrt5$$ or $$64\cdot66>841\cdot5$$ or $$65^2-1>4205$$ or $$4225-1>4205,$$ which is true.

Answer 3

You can always approach it with rational approximations that have "nice" denominators. It's simple to verify the following (see note at bottom):

• $\tfrac{\color{magenta}{22,360}}{10,000} <\sqrt{5} < \tfrac{\color{magenta}{22,361}}{10,000}$
• $\tfrac{\color{red}{24,494}}{10,000} <\sqrt{6} < \tfrac{\color{red}{24,495}}{10,000}$
• $\tfrac {\color{blue}{33,166}}{10,000} <\sqrt{11} < \tfrac {\color{blue}{33,167}}{10,000}$

So we have $$\tfrac{22,360}{10,000} + \tfrac{24,494}{10,000} + \tfrac{33,166}{10,000} < \sqrt{5} + \sqrt{6} + \sqrt{11} < \tfrac{22,361}{10,000} + \tfrac{24,495}{10,000} + \tfrac{33,167}{10,000}$$

$$\tfrac{80,020}{10,000} < \sqrt{5} + \sqrt{6} + \sqrt{11} < \tfrac{80,023}{10,000}$$

In particular, $8 < \sqrt{5} + \sqrt{6} + \sqrt{11}$.

What we really have shown is that $8.0020 < \sqrt{5} + \sqrt{6} + \sqrt{11} <8.0023$, but the desired result follows.

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Note: The initial approximations just come from finding the two perfect squares that bracket $5\times 10^8$, $6\times 10^8$, and $11\times 10^8$. We easily see that

• $\color{magenta}{22,360}^2 = 499,969,600 < 500,000,000 < 500,014,321 = \color{magenta}{22,361}^2$
• $\color{red}{24,494}^2 = 599,956,036 < 600,000,000 < 600,005,025 = \color{red}{24,495}^2$
• $\color{blue}{33,166}^2 = 1,099,983,556 < 1,100,000,000 < 1,100,049,889 = \color{blue}{33,167}^2$

Answer 4

You can reduce the number of radical signs in an inequality by bringing one radical to the other side and squaring. If you do that multiple times you eventially get rid of them all.

If we use the notation: $A <=>B$ to mean $A<B$ or $A=B$ or $A> B$ but we don't know specifically which.

$\sqrt 5 +\sqrt 6 + \sqrt 11 <=> 8$

$\sqrt 5 + \sqrt {11} <=> 8 - \sqrt 6 $.

(Why did I pick $\sqrt 6$ and not $\sqrt 5$ or $11$? No good reason. I just figure $(8-\sqrt 6)^2$ will have more factors and easier to deal with. I could be wrong.)

( Also important to note. We know $8 > \sqrt 6$ so $8-\sqrt 6 > 0$. Both sides have to be positive for this process to work. but if one were positive and the other were negative we would be done because we'd know which side was larger.)

$(\sqrt 5 + \sqrt {11})^2 <=> (8-\sqrt 6)^2$

$5 + 2\sqrt{55}+ 11 <=> 64 - 16\sqrt 6 + 6$

$2\sqrt{55} + 16 <=> 70-16\sqrt 6$

$2\sqrt{55}+16\sqrt {6} <=> 54$

$\sqrt{55} + 8 \sqrt{6} <=>27$

$\sqrt{55} <=> 27-8\sqrt 6$.

($\sqrt 6 < 3$ so $8\sqrt 6 < 24 < 27$ so both sides are positive.)

$(\sqrt{55})^2 <=> (27-8\sqrt 6)^2$

$55 <=> 27^2 - 16\cdot 27\sqrt 6 + 64\cdot 6$

$16\cdot 27\sqrt 6 <=> 27^2 +64\cdot 6 - 55$

(Okay, pencil and paper time. $27^2= (30-3)(30-3) = 900-2*3*30 + 9=900 -180+9=729$ and $729 - 55 = 674$

$16\cdot 27\sqrt 6 <=>674 + 64\cdot 6$

$8\cdot 27\sqrt 6 <=> 337+64\cdot 6$

$(8\cdot 27\sqrt 6)^2 <=> (337 + 64\cdot 6)^2$

$64\cdot 27^2\cdot 6 <=>337^2 + 2\cdot 64\cdot 6+64^2\cdot 36$

$27^2 <=>\frac {337^2}{64\cdot 6} + 2+64\cdot 6$.

$729 <=> \frac{337^2}{360+24} + 2 + (360 + 24)$

$729 - 384-2 <=>\frac {337^2}{384}$

$343 <=> \frac {337^2}{384}$

$343\cdot 384 <=> 337^2$.

But $343 \cdot 384 > 337\cdot 384 > 337\cdot 337 = 337^2$

so the LHS is larger.

So $\sqrt 5 + \sqrt 6 + \sqrt 11 > 8$

Answer 5

The following is a somewhat more laborious, but also more generic, way to solve the problem.

• The minimal (rational) polynomial of $\,x_0=\sqrt{5}+\sqrt{6}+\sqrt{11}\,$ is $$ P(x)=x^8 - 88 x^6 + 1696 x^4 - 10560 x^2 + 14400 $$ which can be determined similar to Minimal Polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$.

• With the substitution $\,x \to x+4\,$: $$P(x+4) = x^8 + 32 x^7 + 360 x^6 + 1472 x^5 - 1504 x^4 - 28160 x^3 - 70976 x^2 - 59904 x - 15296 $$ By Descartes' rule of signs $A(x)=P(x+4)$ has exactly one real positive root, since there is one change of signs and $A(0) \lt 0\,$. It follows that $P(x)$ has exactly one real root larger than $\,4\,$ and that root must be $\,x_0\,$ since $\,x_0=\sqrt{5}+\sqrt{6}+\sqrt{11} \gt \sqrt{4}+\sqrt{4} = 4\,$.

• With the substitution $\,x \to x+8\,$: $$P(x+8) = x^8 + 64 x^7 + 1704 x^6 + 24448 x^5 + 203936 x^4 + 988160 x^3 + 2574016 x^2 + 2780160 x - 6080 $$ Again by Descartes' rule of signs $B(x)=P(x+8)$ has exactly one real positive root, since there is one change of signs and $B(0) \lt 0\,$. It follows that $P(x)$ has exactly one real root larger than $\,8\,$, and that root must be the unique root larger than $\,4\,$ as established at the previous step, so in the end $\,x_0 \gt 8\,$.

As a side note, the middle step and related calculations could have been avoided by noting that the roots of $\,P(x)\,$ are the rational conjugates $\,\pm\sqrt{5}\pm\sqrt{6}\pm\sqrt{11}\,$, so $\,x_0\,$ is the largest real root.

Answer 6

Use the Taylor expansions: $$\sqrt{1+x}=\sqrt{x} + \frac1{2\sqrt x} - \frac1{8x\sqrt x}+\frac1{16x^2\sqrt x}-\frac5{128x^3\sqrt x}+O(x^{-4})\\ \sqrt{2+x}=\sqrt{x} + \frac1{\sqrt x} - \frac1{2x\sqrt x}+\frac1{2x^2\sqrt x}-\frac5{8x^3\sqrt x}+O(x^{-4})$$ Now we will plug in and calculate (manually possible, though tedious): $$2.23606...=\sqrt{1+4}>2+\frac1{4}-\frac1{64}+\frac1{512}-\frac5{16384}\approx 2.23602;\\ 2.4494...=\sqrt{2+4}>2+\frac12-\frac1{16}+\frac1{64}-\frac5{1024}\approx 2.4482;\\ 3.3166...=\sqrt{2+9}>3+\frac13-\frac1{54}+\frac1{486}-\frac5{17496}\approx 3.3165.$$ Hence: $$\sqrt5+\sqrt6+\sqrt{11}>8.00062>8$$