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: 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
