Proving $\sqrt{100,001}-\sqrt{100,000} < \frac{1}{2\sqrt{100,000}}$

I squared both sides of the equation to get

$100,001 + 100,000+-200\sqrt{10}\sqrt{100,001} < \frac{1}{400,000}$.

I am just not sure how to justify this. I've tried multiplying both sides by -1, but it still would not hold.


$$\sqrt{a} - \sqrt{b} = \frac{a-b}{\sqrt{a} + \sqrt{b}}$$

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    $\begingroup$ I don't follow. So would we change the left hand side of the equation to 1/((sqrt(100,001)-sqrt(100,000))? I don't see how this makes sense because the denominator would be extremely small which would make the left hand side of the equation greater than the right. $\endgroup$ – user3699546 Jan 30 '15 at 14:45
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    $\begingroup$ @user3699546 I answered 17 minutes ago. Give yourself maybe more time for thinking about it, and it will maybe make sense. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 30 '15 at 14:47
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    $\begingroup$ Oh, okay I think I got it. I have to justify that the sqrt(100,000)+sqrt(100,001) is greater than 2*sqrt(100000), which would mean that the right hand side is indeed larger. $\endgroup$ – user3699546 Jan 30 '15 at 15:14
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    $\begingroup$ correct ! you're right $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 30 '15 at 15:14
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    $\begingroup$ It is derived from $(X-Y)(X+Y) = X^2 - Y^2$, with $X = \sqrt{a}$ and $Y = \sqrt{b}$. In french these kinds of equalities are called "remarkable identities/equalities." You have others : $(X+Y)^2 = X^2 + 2XY + Y^2$, etc etc. If you're happy with my answer, do not hesitate to validate it; so that the "case" is closed. ;-) $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Jan 30 '15 at 15:18

Another way:

$$\sqrt{a+1}-\sqrt a<\frac1{2\sqrt a}.$$ Multiplying by $2\sqrt a$ and moving the terms,




This works for any $a>0$ !


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