# Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$

Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

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You are asing to show that $\sqrt{x}$ is concave. – gt6989b May 17 '13 at 16:10

$$\frac{1}{\sqrt{1000}+\sqrt{1001}}<\frac{1}{\sqrt{1000}+\sqrt{999}}$$

$$\implies \sqrt{1001}-\sqrt{1000}<\sqrt{1000}-\sqrt{999}$$

$$\implies \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$$

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Nice. (First step is $\frac1{\sqrt{a}+\sqrt{b}} = \frac{\sqrt{a}-\sqrt{b}}{a-b}$ where $a-b=1$.) – Myself May 17 '13 at 16:16
Simplest of the lot, $(+1)$ – lab bhattacharjee May 17 '13 at 16:26
+1 nice one from a 12th grade student(i guess CBSE?) – iostream007 May 17 '13 at 17:09
@iostream007:ISC actually. – shaswata May 17 '13 at 17:12
@shaswata I think today was result declare? – iostream007 May 17 '13 at 17:40

This is an overkill solution, but this is an immediate application of Cauchy-Schwarz:

$$\left( 1 \cdot \sqrt{1001}+1 \cdot\sqrt{999}\right)^2 \leq (1+1)(1001+999)=4000\,.$$

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Just what I was thinking! – Soham Chowdhury May 18 '13 at 4:35

The answer is: YES, we can!

\begin{align} (\sqrt{1001}+\sqrt{999})^2&=2000+2\sqrt{1001\times 999} \\ &=2000+2\sqrt{(1000+1)(1000-1)} \\ &=2000+2\sqrt{1000^2-1} \end{align} \begin{align} \text{and that: }(2\sqrt{1000})^2&=4000 \\ &=2000+2\sqrt{1000^2} \end{align}

Since $2000+2\sqrt{1000^2-1}<2000+2\sqrt{1000^2}$

$\therefore \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$

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Isn't the second to last inequality is backwards? – Student May 17 '13 at 16:14
Well, yes, I was trying to make it a little bit easier to understand – user67258 May 17 '13 at 16:16
@DannyCheuk I think it's still backwards... if $2000 + 2\sqrt{1000^2 - 1} > 2000 + 2\sqrt{1000^2}$ then it must follow that $1000^2 - 1 > 1000^2$ and so $-1 > 0$? – DanZimm May 17 '13 at 16:38

You can tell without calculation if you can visualize the graph of the square-root function; specifically, you need to know that the graph is concave (i.e., it opens downward). Imagine the part of the graph of $y=\sqrt x$ where $x$ ranges from $999$ to $1001$. $\sqrt{1000}$ is the $y$-coordinate of the point on the graph directly above the midpoint, $1000$, of that interval. $\frac12(\sqrt{999}+\sqrt{1001})$ is the average of the $y$-coordinates at the ends of this segment of the graph, so it's the $y$-coordinate of the point directly above $x=1000$ on the chord of the graph joining those two ends. The concavity of the graph shows that the chord lies below the graph. So $\frac12(\sqrt{999}+\sqrt{1001})<\sqrt{1000}$. Multiply by $2$ to get the numbers in your question.

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If you could embed a graphic, this would be a great answer... – Alex Feinman May 17 '13 at 17:04
@AlexFeinman: It would be nice. Unfortunately, the square root function is so close to linear over this range (which is why the calculator values are so close) that the difference between the two curves does not show up. – Ross Millikan May 17 '13 at 17:51
Perhaps exaggerate the effect by choosing smaller values? – Sean Allred May 18 '13 at 4:36

From Jensen's inequality, the mean of the square root is less than the square root of the mean, unless the numbers are equal. This is true for more than two numbers as well.

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IMO, the most obvious first thing to do when comparing $\sqrt{999} + \sqrt{1001}$ with $2 \sqrt{1000}$ is to split the right hand side into a sum of two copies of $\sqrt{1000}$ and compare $\sqrt{1000}$ with $\sqrt{999}$ and $\sqrt{1000}$ with $\sqrt{1001}$.

By looking at the shape of the graph of $\sqrt{x}$, we can see that it grows faster with smaller numbers. The difference between $\sqrt{999}$ and $\sqrt{1000}$ should be slightly larger than the difference between $\sqrt{1000}$ and $\sqrt{1001}$. Thus, we get a greater loss from replacing one copy of $\sqrt{1000}$ with $\sqrt{999}$ than the gain we get by replacing the other copy of $\sqrt{1000}$ with $\sqrt{1001}$, and so should expect

$$\sqrt{999} + \sqrt{1001} < 2\sqrt{1000}$$

Okay, but how can we be sure that our reasoning is actually correct? We may have overlooked something, or our intuition might be flat out wrong! How can we turn this argument into a rigorous one?

Normally, this would be a wonderful job for calculus; the main idea of the proof is very closely related to the fact that the second derivative of $f(x) = \sqrt{x}$ is negative. But since you tagged this precalculus, let's find another way....

There are a number of algebraic tricks for doing this sort of thing without the help of calculus, e.g. by manipulating the problem to eliminate square roots. Ad-hoc methods often work well for simple problems, as in the other answers.

But what if we don't find something ad hoc? Here is something more direct.

In order to understand $\sqrt{1000} - \sqrt{999}$, we can multiply by it its conjugate to simplify (and divide it out so as not to change the overall value)

\begin{align} \sqrt{1000} - \sqrt{999} &= \frac{(\sqrt{1000} - \sqrt{999})(\sqrt{1000} + \sqrt{999})}{\sqrt{1000} + \sqrt{999}} \\&= \frac{1000 - 999}{\sqrt{1000} + \sqrt{999}} \\&= \frac{1}{\sqrt{1000} + \sqrt{999}} \end{align}

Multiplying by conjugates is very commonly useful. While we still might not be able to fully understand $\sqrt{1000} + \sqrt{999}$, it is much easier to understand than $\sqrt{1000} - \sqrt{999}$, because its most significant parts combine constructively rather than cancel each other out. This tells us the difference is very nearly

$$\frac{1}{2 \sqrt{1000}}$$

We'll get a similar result by applying the same method to $\sqrt{1001} - \sqrt{1000}$ (in fact, we can use the exact forms of the result directly, but let's assume we didn't notice that). So we need to go a step further and figure out how much the true difference differs from this estimate:

\begin{align} \left(\sqrt{1000}- \sqrt{999} \right) - \frac{1}{2 \sqrt{1000}} &= \frac{1}{\sqrt{1000} + \sqrt{999}} - \frac{1}{2 \sqrt{1000}} \\&= \frac{2 \sqrt{1000} - \sqrt{1000} - \sqrt{999}}{2 \sqrt{1000} (\sqrt{1000} + \sqrt{999})} \\&= \frac{\sqrt{1000} - \sqrt{999}}{2 \sqrt{1000} (\sqrt{1000} + \sqrt{999})} \\&= \frac{1}{2 \sqrt{1000} (\sqrt{1000} + \sqrt{999})^2} \end{align}

Normally I would have had to use the "multiply by conjugate" trick in the numerator again, but we've already seen how to deal with $\sqrt{1000} - \sqrt{999}$ so I could just plug in what we already knew.

Similarly

$$\left(\sqrt{1001}- \sqrt{1000} \right) - \frac{1}{2 \sqrt{1000}} = -\frac{1}{2 \sqrt{1000} (\sqrt{1001} + \sqrt{1000})^2}$$

And now the two numbers are easily distinguishable: one is positive and one is negative!

And so

$$\left(\sqrt{1000}- \sqrt{999} \right) - \frac{1}{2 \sqrt{1000}} > \left(\sqrt{1001}- \sqrt{1000} \right) - \frac{1}{2 \sqrt{1000}}$$

which can be simplified to give the desired answer.

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$\sqrt{1001}+\sqrt{1000}>\sqrt{1000}+\sqrt{999} \\ \implies \dfrac{1}{\sqrt{1001}-\sqrt{1000}}>\dfrac{1}{\sqrt{1000}-\sqrt{999}}\\ \implies\sqrt{1000}-\sqrt{999}>\sqrt{1001}-\sqrt{1000} \\ \implies 2\sqrt{1000}>\sqrt{999}+\sqrt{1001}$

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some one already give same answer btw +1 for your effort – iostream007 May 17 '13 at 17:10
@iostream007: I post my answer and then notify that. – Argha May 17 '13 at 18:04

$\sqrt x$ is a concave function, therefore $\alpha \sqrt x + (1-\alpha) \sqrt y < \sqrt {x+y }.$ Substitute $\alpha = \frac12, x=999, \text{and } y=1001$

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And one more... I'm reflecting at the concavity and monotonicty of the curve of the sqrt-function and conclude, that whatever greater or smaller the relation is, it should be the same if I decrease the arguments down to zero. Since we do not yet know whether the relation is $\gt$ or $\lt$ I introduce the "indeterminate comparision symbol" $\mathcal C$ which can assume the greater or smaller-relation and we rewrite the original question $$\sqrt{1001} + \sqrt{ 999} \qquad \mathcal C \qquad 2\sqrt{1000} \tag 1$$ as $$\sqrt{1001} - \sqrt{1000} \qquad \mathcal C \qquad \sqrt{1000} - \sqrt{999} \tag 2$$ Then we use the monotonicity and concavity of the sqrt-function (from x=1001 down to x=0) and write the consecutive differences

$\displaystyle \qquad \begin{array} {lll} &\sqrt{1001} & - & \sqrt{1000} & \mathcal C & \sqrt{1000} &-& \sqrt{999} \\ &\sqrt{1000} & - & \sqrt{ 999} & \mathcal C& \sqrt{ 999} &-& \sqrt{998} \\ &\sqrt{ 999} & - & \sqrt{ 998} & \mathcal C& \sqrt{ 998} &-& \sqrt{997} \\ &\cdots \\ &\sqrt{ 2} & - & \sqrt{ 1} & \mathcal C & \sqrt{ 1} &-& \sqrt{ 0} \\ \end{array}$

and compare the whole sums which are nicely telescoping

$\displaystyle \qquad \begin{array} {lll} \sum =&\sqrt{1001} &-&1 & \mathcal C & \sqrt{1000} &&& \end{array}$ $\tag 3$

which can then be rewritten by

$\displaystyle \qquad \begin{array} {rrrrrrr} &\sqrt{1001} &-&\sqrt{1000} & \mathcal C & 1 && & // * (\sqrt{1001} + \sqrt{1000})\\ & 1001 & - & 1000 & \mathcal C & \sqrt{1001} &+& \sqrt{1000} \\ & & & 1 & \mathcal C & \sqrt{1001} &+& \sqrt{1000} &\sim 2\sqrt{1000} \\ \end{array}$

Here, in the last comparision, the geater/smaller-relation is obvious and thus our operator $\mathcal C = "\lt"$ and we have the result

$\displaystyle \qquad \begin{array} {lll} &\sqrt{1001} & + & \sqrt{ 999} & \lt & 2 \cdot \sqrt{1000} \end{array}$ $\tag 4$

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Square, subtract 2000, divide by 2 for both sides, square again and end up with 1001*999 vs 1000*1000 that is (1000+1)(1000-1) vs 1000*1000 or 1000 000 -1 vs 1000 000

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Almost everyone $\bf guesses$ a reasonable answer and then verifies it. However, this is a rather calculational field of mathematics and so the result should be $\bf calculated$.

Let $\mathcal{X}$ be one of the comparions symbols $\leq,=,\geq$. Let us solve for $\mathcal{X}$ in the formula $\sqrt{1001}+\sqrt{999}\ \;\;\;\mathcal{X}\;\;\; \ 2\sqrt{1000}$.

Let us calculate,

$\hspace{0.5cm}\sqrt{1001}+\sqrt{999}\ \;\;\;\mathcal{X}\;\;\; \ 2\sqrt{1000} \\ \equiv \;\; (\sqrt{1001}+\sqrt{999})^2 \;\;\;\mathcal{X}\;\;\; (2\sqrt{1000})^2 \hspace{1.3cm}\text{,$\mathcal{X}$invariant under squaring/ positive-square roots}\\ \equiv \;\; 1001+2\sqrt{1001}\sqrt{999} + 999 \;\;\;\mathcal{X}\;\;\; 4\cdot 1000 \hspace{0.5cm}\text{,arithmetic}\\ \equiv \;\; 2\sqrt{1001}\sqrt{999} \;\;\;\mathcal{X}\;\;\; 2000 \hspace{2.9cm}\text{,$\mathcal{X}$invariant under addition/subtraction; arithmetic}\\ \equiv \;\; 4 \cdot 1001 \cdot 999 \;\;\;\mathcal{X}\;\;\; 2000^2 \hspace{2.5cm}\text{,$\mathcal{X}$invariant under squaring/ positive-square roots}\\ \equiv \;\; 3999996 \;\;\;\mathcal{X}\;\;\; 4000000 \hspace{3.0cm}\text{,arithmetic}\\ \equiv \;\; 0 \;\;\;\mathcal{X}\;\;\; 4 \hspace{5.2cm}\text{,$\mathcal{X}$invariant under addition/subtraction; arithmetic}\\ \equiv \;\; \mathcal{X} \;\text{ is }\le" \hspace{5cm}\text{,possibilities for$\mathcal{X}$, 0 is at-most 4}\\$

Thus, using properties about the comparisons operators, we have $\bf solved$ for the relation between the given quantities; there was never any guess nor any $\textbf{rabbit-out-of-a-hat}$ moves in the above calculation. It is guided by the need to simplify.

Note0 :: This process I first witnessed in a text by Roland Backhouse on Program Specification and Refinement.

Note1 :: User Gottfried Helms'' gives similar approach but using different properties of the comparison operators. http://math.stackexchange.com/a/400390/80406.

Best regards,

Moses

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This is connected to some of the other answers, but I thought it might be worth mentioning to show the utility of approximation formulas. The first two derivatives of $\ f(x) = x^{1/2} \$ at $\ a = 1000 \$ are $\ f'(1000) = \frac{1}{2} \cdot 1000^{-1/2} \$ and $\ f''(1000) = -\frac{1}{4} \cdot 1000^{-3/2} \$ (we won't need these to be evaluated). The Taylor series for $\ x^{1/2} \$ about $\ a = 1000 \$ is then

$$f(x) \ = \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (x - 1000) \ - \ \frac{1}{4\cdot 1000^{3/2}} (x-1000)^2 \ + \ \ldots$$

Hence,

$$999^{1/2} \ + \ 1001^{1/2} \ \approx$$

$$[ \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (-1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (-1)^2 \ + \ \ldots$$

$$+ \ 1000^{1/2} \ + \ \frac{1}{2\cdot 1000^{1/2}} (1) \ - \ \frac{1}{4\cdot 1000^{3/2}} (1)^2 \ + \ \ldots \ ]$$

$$= \ 2 \cdot 1000^{1/2} \ - \ \frac{2}{4\cdot 1000^{3/2}} \ - \ \ldots \ ,$$

with all the terms in odd powers of $\ (x-1000) \$ cancelling and all of the higher-order terms in even powers also being negative. Because the two square-roots are being added, it is not sufficient to only carry this calculation to linear terms ("first-order") ; this is hinted at by the fact that OP found by calculator how close the two values are.

ADDED: The linear terms are sufficient, though, to show that $\ \sqrt{1001} \cdot \sqrt{999} < 1000 \ .$

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