Coincidental decimal representation?

We have that $D = (\sqrt{320} - 17) - (\sqrt{79} - 8) = 2\sqrt{80} - \sqrt{79} - 9 = 2\sqrt{80} - (\sqrt{79} + \sqrt{81})$ .

Now $(\sqrt{79} + \sqrt{81})^{2} = 79 + 81 + 2\sqrt{79 \times 81} = 160 + 2\sqrt{(80 - 1)(80 + 1)} =$

$160 + 2\sqrt{80^{2} - 1} = 160 + 160\sqrt{1 - \dfrac{1}{6400}} \lt 160 + 160 = 320$ .

Thus $\sqrt{79} + \sqrt{81} \lt \sqrt{320} = 2\sqrt{80}$ , and so $D \gt 0$ , i.e., $\boxed{\sqrt{320} - 17 \gt \sqrt{79} - 8}$ .

Starting with:

$\sqrt{79}-8 ><= \sqrt{320}-17$

We can simplify to:

$\sqrt{79}-8 ><= \sqrt{320}-9-8$

$\sqrt{79} ><= \sqrt{320}-9$

$\sqrt{79} ><= \sqrt{320}-\sqrt{81}$

$\sqrt{79} ><= 2\sqrt{80}-\sqrt{81}$

Generalizing to n:

$\sqrt{n-1} ><= 2\sqrt{n}-\sqrt{n+1}$

$\sqrt{n-1}+\sqrt{n+1} ><= 2\sqrt{n}$

$(\sqrt{n-1}+\sqrt{n+1})^2><= (2\sqrt{n})^2$

$n-1 + 2(\sqrt{n-1}\sqrt{n+1}) +n+1><= 4n$

$2n+2(\sqrt{n-1}\sqrt{n+1})><= 4n$

$n+\sqrt{n-1}\sqrt{n+1}><= 2n$

$\sqrt{n-1}\sqrt{n+1}><= n$

$\sqrt{(n-1)(n+1)} ><= n$

$(n-1)(n+1) ><= n^2$

$n^2-1 ><= n^2$

Therefore the sign of the inequality is < for all positive real numbers and:

$\sqrt{n-1} < 2\sqrt{n}-\sqrt{n+1}$

$\begin{aligned}\text{Let ,}D&=(\sqrt{320}-17)-(\sqrt{79}-8)\\ &=2\sqrt{80}-\sqrt{79}-9\\ &=(\sqrt{80}-\sqrt{79})+(\sqrt{80}-\sqrt{81})\\\\ \text{Let,} f(x)&=\sqrt{x+1}-\sqrt{x}\\\\ f'(x)&=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{x+1}}-\dfrac{1}{\sqrt{x}}\right)\\\\ \sqrt{x+1}&>\sqrt{x} \text{ for positive reals}\\\\ \implies f'(x)&<0 \text{ for all }x\geq0\\\\ f(x) & \text{ is strictly decreasing}\\\\ \implies\sqrt{80}&-\sqrt{79}>\sqrt{81}-\sqrt{80}\\\\ \text{ or }(\sqrt{80}&-\sqrt{79})+(\sqrt{80}-\sqrt{81})>0\\\\ \implies D&>0\\\\ \text{ Thus, }&(\sqrt{320}-17)>(\sqrt{79}-8)\\\\\\ \text{Alternate}&\text{ solution:}\\\\ D&=(\sqrt{80}-\sqrt{79})+(\sqrt{80}-\sqrt{81})\\\\ \text{Rationalizing}&\text{ the numerator, we get}\\\\ D&=(\sqrt{80}-\sqrt{79})\dfrac{(\sqrt{80}+\sqrt{79})}{(\sqrt{80}+\sqrt{79})}-(\sqrt{81}-\sqrt{80})\dfrac{(\sqrt{80}+\sqrt{81})}{(\sqrt{80}+\sqrt{81})}\\\\ D&=\dfrac{1}{(\sqrt{80}+\sqrt{79})}-\dfrac{1}{(\sqrt{80}+\sqrt{81})}\\\\ &(\sqrt{80}+\sqrt{79})<(\sqrt{80}+\sqrt{81})\\\\ \implies& \dfrac{1}{(\sqrt{80}+\sqrt{79})}>\dfrac{1}{(\sqrt{80}+\sqrt{81})}\\\\ \implies D&>0\\\\ \text{ Thus, }&(\sqrt{320}-17)>(\sqrt{79}-8)\\\\\\\end{aligned}$

I worked out the repeating continued fraction expansions of both numbers: $\sqrt{320} - 17 = \cfrac 1 {1 + \cfrac 1 {7 + \cfrac 1 {1 + \cfrac 1 {34 + \ddots}}}}$ The sequence $1, 7, 1, 34$ repeats. The other number, $\sqrt{79}-8$ had $1,7,1,16$ repeating. One may deduce the answer from that.

Postscript in response to comments: I DID NOT use the decimal expansion to find the continued fraction expansion. Since we know $17^2 < 320 < 18^2$ , you have $\sqrt{320} - 17 = \dfrac 1 {\text{something} > 1}$ , and then you say $\sqrt{320} - 17 = \dfrac{320 - 17^2}{\sqrt{320} + 17} = \dfrac{31}{\sqrt{320}+17}$ (this is rationalizing the numerator) and that is equal to $\dfrac 1 {\left(\frac{\sqrt{320} + 17}{31}\right)}$ . Next, we have $\sqrt{320} + 17$ between $34$ and $35$ , so dividing it by $31$ we get something between $1$ and $2$ , so we have $\cfrac 1 {1 + \cfrac{\sqrt{320} - 14}{31}}$ . Now rationalize the numerator again, getting $\cfrac 1 {1 + \cfrac 4 {\sqrt{320} + 14} }$ . Keep going like that until you see the same expression you saw at an earlier step, and then you know it has to repeat.

Relevant wiki: Jensen's Inequality

Consider the function $f(x) = \sqrt{81-x} - (9-\tfrac12 x).$ Clearly, $f(0) = 0$ . Now $f''(x) = \frac{-1}{4(81-x)^{3/2}} < 0,$ showing that the function is convex on the interval $0 \leq x \leq 2$ . Jensen's inequality states that $f\left(\tfrac12(0+2)\right) \geq \tfrac12\left(f(0) + f(2)\right),$ i.e. $2f(1) \geq f(2).$ Therefore $2(\sqrt{80} - 8\tfrac12) \geq \sqrt{79} - 8,$ that is, $\boxed{\sqrt{320} - 17} \geq \sqrt{79} - 8.$

People are really bringing out the big guns for this problem. This can be done with simple algebra.

Generally, when comparing the numbers $\sqrt{a} + b$ and $\sqrt{c} + d$ , remember that addition and subtraction preserve order, as does squaring (as long as both sides of the inequality are positive). The simple thing to do is rearrange terms so the integers are on one side of the inequality while the radicals are on the other side. Then square. Sort terms one more time, square again, and look at the final inequality.

In this case $\sqrt{79} - 8 < \sqrt{320} - 17 \Leftrightarrow$

$9 < \sqrt{320} - \sqrt{79} \Leftrightarrow$

$81 < 399 -2*\sqrt{320*79} \Leftrightarrow$

$2 * \sqrt{320*79} < 318 \Leftrightarrow$

$4 * 320 * 79 < 318*318 \Leftrightarrow$

$101120 < 101124$ Since the last inequality is true, the first one is also true.

The answer is ✓(320)-17

Here's how I rationalize this WITHOUT using a calculator. ...

If:

✓(320)=2×✓(80)

          &

 -17= 2×(-8.5)

And we also know that:

✓(80) > ✓(79)

       and: 

  -8  >  -8.5

Then it would stand to reason that:

2×✓(80) > ✓(79)

       &

 -8 > 2(-8.5)

Which also means that:

2×[✓(80)-8.5] > ✓(79)-8

And because :

2×[✓(80)-8.5]= ✓(320)-17

We can conclude that:

✓(320)-17> ✓(79)-8

Therefore the answer is:

✓(320)-17

Square both expressions so that the difference increases ☺

We will first conjecture that sqrt(320) - 17 > sqrt(79) - 8 and set out to apply inequality algebra to prove were right.
1) Add +17 to both sides: sqrt(320) > sqrt(79) + 9.
2) Square both sides: 320 > 79 + 18 sqrt(79) + 81,
320 > 160 + 18 sqrt(79)
160 > 18 * sqrt(79)

3) Square to check:
25,600 > (18^2) * 79
25,600 > 324 * 79
25,600 > 25,596. check.

Henceforth, sqrt(320) - 17 > sqrt(79) - 8. QED.

These numbers $79$ and $320$ are not perfect squares. The roots of the previous perfect square are subtracted from the root of each number. the root. So, both the two expression will obviously yield 0 point something. The further away they are from the previous perfect square, the greater their roots. Since $320-17^2=31$ is greater than $79-8^2=15$ , the answer is the latter option.

It never occurred to me to do it /without/ a calculator...