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I am stuck in a difficult question:

Compare $18$ and $$ A=\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40} $$ without using calculator.

Thank you for all solution.

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  • $\begingroup$ There is no question here $\endgroup$
    – user223391
    Dec 19 '17 at 16:11
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    $\begingroup$ Try bounding the sums $\sqrt{7}+\sqrt{11}$ and $\sqrt{32}+\sqrt{40}$ $\endgroup$
    – kingW3
    Dec 19 '17 at 16:15
  • $\begingroup$ A first try is to bound $\sqrt 7 \lt \sqrt 8$ and $\sqrt {11} \lt \sqrt {12}$. Then the sum becomes $2\sqrt 2+2\sqrt 3 + 4\sqrt 2 + 2\sqrt{10}$. If you plug in $1.4,1.7,3.1$ the sum becomes $18$ and all of these are below the square roots, so we need a pretty fine approximation. It doesn't work, but this is one thing to try. $\endgroup$ Dec 19 '17 at 16:20
  • $\begingroup$ Generally, you will get close votes if you don't display something, even if it is wrong, that indicates that you have tried to solve the problem on your own. $\endgroup$ Dec 19 '17 at 17:33
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From the guidance of kingW3 I give a complete solution. We will prove that $$ \sqrt{7}+\sqrt{11}<6, \quad \sqrt{32}+\sqrt{40}<12 $$ by taking squaring both sides of inequalities. Indeed, we have $$ \sqrt{7}+\sqrt{11}<6\Leftrightarrow 7+11+2\sqrt{77}<36\Leftrightarrow\sqrt{77}<9, $$ $$ \sqrt{32}+\sqrt{40}<12\Leftrightarrow 32+40+2\sqrt{1280}<144\Leftrightarrow\sqrt{1280}<36. $$

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  • $\begingroup$ You mean $\sqrt {11}$ there. $\endgroup$ Dec 19 '17 at 16:22
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    $\begingroup$ Nice one (+1), the intuition I had behind this was that $7,11$ can be written as $9-2,9+2$ and for any $a$ $\sqrt{a-x}+\sqrt{a+x}\leq 2\sqrt a$ similarly with $32,40$ as $36-4,36+4$. $\endgroup$
    – kingW3
    Dec 19 '17 at 16:47
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$\sqrt{7}$ and $\sqrt{11}$ can be written as $\sqrt{9\pm 2}$ and similarly $\sqrt{32}$ and $\sqrt{40}$ can be written as $\sqrt{36\pm 4}$.
Since $\sqrt{x}$ is a concave function on $\mathbb{R}^+$,

$$ \sqrt{9-2}+\sqrt{9+2}+\sqrt{36-4}+\sqrt{36+4} \color{red}{\leq} 2\sqrt{9}+2\sqrt{36} = 18.$$ We may also estimate the difference between the RHS and the LHS: $$\begin{eqnarray*} 2n-\sqrt{n^2+x}-\sqrt{n^2-x}&=&\frac{x}{n+\sqrt{n^2-x}}-\frac{x}{n+\sqrt{n^2+x}}\\&=&x\cdot \frac{\sqrt{n^2+x}-\sqrt{n^2-x}}{(n+\sqrt{n^2-x})(n+\sqrt{n^2+x})}\\&=&\frac{2x^2}{(n+\sqrt{n^2-x})(n+\sqrt{n^2+x})(\sqrt{n^2+x}+\sqrt{n^2-x})}\\&\geq&\frac{x^2}{n(3n^2+\sqrt{n^4-x^2})}\geq\frac{x^2}{n\left(4n^2-\frac{x^2}{2n^2}\right)}.\end{eqnarray*} $$ This leads to $18-\left(\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}\right)\geq \frac{1}{18}$.

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$$\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}=\sqrt{7+40+2\sqrt{7\cdot40}}+\sqrt{11+32+2\sqrt{11\cdot32}}=$$ $$=\sqrt{47+2\sqrt{280}}+\sqrt{43+2\sqrt{352}}<\sqrt{47+2\cdot17}+\sqrt{43+2\cdot19}=18$$

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$$\begin{align} 3\left(\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}\right) &= \sqrt{63}+\sqrt{99}+\sqrt{288}+\sqrt{360}\\ &< \sqrt{64}+\sqrt{100}+\sqrt{289}+\sqrt{361}\\ &< 8+10+17+19\\ &< 3\times 18\end{align}$$

We can even estimate from $\sqrt{1+x}$ expansion that the error is about

$\epsilon=-\frac 16(\frac 18+\frac 1{10}+\frac 1{17}+\frac 1{19})\approx -0.056\quad$ giving $\quad \approx 17.944$

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Lets be methodical about this. Consider finding a nice upper bound for $\sqrt a + \sqrt b$.

$$(\sqrt a + \sqrt b)^2 = a + b + 2\sqrt{ab}$$

So maybe we should replace $ab$ with the smallest $n$ such that $ab \le n^2$

Then $(\sqrt a + \sqrt b)^2 < a + b + 2n$. So

$$\sqrt a + \sqrt b < \sqrt{a + b + 2n}$$


We note that $7 \times 32 = 224 < 225 = 15^2$ and $11 \times 40 = 440 < 441 = 21^2$

Then

$$\sqrt 7 + \sqrt{32} < \sqrt{7 + 32 + 2\times 15} = \sqrt{69}$$

And

$$\sqrt{11} + \sqrt{40} < \sqrt{11 + 40 + 2\times 21} = \sqrt{93}$$

So $$\sqrt{69} + \sqrt{93} < \sqrt{69 + 93 + 2 \times 81} = 18$$

Hence $$\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40} < 18$$

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