# Compare sum of radicals

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.

• There is no question here
– user223391
Dec 19 '17 at 16:11
• Try bounding the sums $\sqrt{7}+\sqrt{11}$ and $\sqrt{32}+\sqrt{40}$ Dec 19 '17 at 16:15
• 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. Dec 19 '17 at 16:20
• 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. Dec 19 '17 at 17:33

## 5 Answers

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 $$\sqrt{7}+\sqrt{11}<6\Leftrightarrow 7+11+2\sqrt{77}<36\Leftrightarrow\sqrt{77}<9,$$ $$\sqrt{32}+\sqrt{40}<12\Leftrightarrow 32+40+2\sqrt{1280}<144\Leftrightarrow\sqrt{1280}<36.$$

• You mean $\sqrt {11}$ there. Dec 19 '17 at 16:22
• 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 $36-4,36+4$. Dec 19 '17 at 16:47

$\sqrt{7}$ 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 $18-\left(\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}\right)\geq \frac{1}{18}$.

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

\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 $\quad \approx 17.944$

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