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.

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.
