Prove the inequality $\,\frac{1}{\sqrt{1}+ \sqrt{3}} +\frac{1}{\sqrt{5}+ \sqrt{7} }+\ldots+\frac{1}{\sqrt{9997}+\sqrt{9999}}\gt 24$ 
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.
 A: Hint: Telescope your rationalized sum by adding terms.
A: 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.
A: Note that $\frac{\sqrt{3}-1}{2}>\frac{\sqrt{5}-\sqrt{3}}{2}$, etc. because $\sqrt{\phantom{x}}$ is concave down. So twice your left-hand side is greater than a telescoping sum.
A: 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
