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Prove the inequality $$\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+......... +\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$

My work:

Rationalizing the denominator gives

$$\frac{\sqrt{3}-1}{2} +\frac{\sqrt{7}-\sqrt{5}}{2}+......+\frac{\sqrt{9999}-\sqrt{9997}}{2} .$$

Now by taking two as common and separating the positive and negative terms gives

$$\frac{1}{2} [ \{\sqrt{3} +\sqrt{7}+\dots +\sqrt{9999}\} - \{1+\sqrt{5} +\dots+\sqrt{9997}\}].$$

Can we do like this please suggest. Thanks.

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HINT: You need to add and subtract somethings to make it a telescope sum. –  some1.new4u Aug 7 '13 at 17:48
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4 Answers

up vote 3 down vote accepted

Another idea: by considering concavity and a left endpoint approximation, the desired sum is an overestimate of the following integral:

$$\frac{1}{2}\int_0^{2500}\sqrt{4x+3}-\sqrt{4x+1}\approx 24.6528$$

More explicitly, notice that your sum is:

$$\frac{1}{2}\sum_{n=0}^{2499}\sqrt{4n+3}-\sqrt{4n+1}$$

We can think of this as one half of the sum of the areas of $2500$ rectangles of width $1$ and height $\sqrt{4n+3}-\sqrt{4n+1}$. These rectangles can be visualized in the plane as follows: consider the two curves $f(x)=\sqrt{4x+3}$ and $g(x)=\sqrt{4x+1}$. The $n$th rectangle (starting the count from $0$) is then formed by the $4$ points:

$$(n,f(n)),(n,g(n)),(n+1,f(n)),(n+1,g(n))$$

Notice that the base has length $1$, and the height is exactly $\sqrt{4n+3}-\sqrt{4n+1}$. Also, notice that the area of the rectangle is well-approximated by the area between the two curves, and in fact is an overestimate if you consider the fact that the upper curve always has a shallower slope. This means that the area between these two curves from $x=0$ to $x=2500$ is an underestimate of your desired sum. This is what the integral above calculates, the area between the two curves.

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This is nice and I'd upvote it if you'd explain the OP what partition of that interval to choose and how to choose the points in each subinterval. After all, the question's tagged "precalculus" and thus a little explanation's due for the answer. –  DonAntonio Aug 7 '13 at 18:25
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@DonAntonio: You're right. I didn't notice the tags. –  Jared Aug 7 '13 at 18:45
    
+1......................... –  DonAntonio Aug 7 '13 at 18:49
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Note that $\frac{\sqrt{3}-1}{2}>\frac{\sqrt{5}-\sqrt{3}}{2}$, etc. So twice your left-hand side is greater than a telescoping sum.

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Hint: Telescope your rationalized sum by adding terms.

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Another idea: Using the inequality $$\frac{1}{\sqrt{2n-1} +\sqrt{2n+1}}\gt\frac{1}{2\sqrt{2n}}, n\ge1$$ we get the folliwng chain of equities/inequalities: $$\sum_{i = 1}^{4999}{\frac{1}{\sqrt{2i-1} +\sqrt{2i+1}}} \gt \sum_{i = 1}^{4999}{\frac{1}{2\sqrt{2i}}} = \frac{1}{2\sqrt{2}}\sum_{i = 1}^{4999}{\frac{1}{\sqrt{i}}} \ge \frac{1}{2\sqrt{2}}\frac{4999}{\sqrt{\frac{\sum_{i = 1}^{4999}{i}}{4999}}} = \frac{1}{2\sqrt{2}} \frac{4999}{\sqrt{2500}} \approx 35.35$$ The last inequality is obtained by using the Root mean square-Harmonic mean inequality

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Cool, but I don't think you are allowed to use those inequalities in pre-calculus (Cauchy-Schwartz) –  Emanuel Landeholm Aug 8 '13 at 0:38
    
I've used them in high school so I don't see a problem and besides all those inequalities have proofs using only elementary mathematics –  lnwvr Aug 8 '13 at 8:07
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