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$\displaystyle \sum_{k=0}^{2499} \frac{1}{\sqrt{4k+1}+\sqrt{4k+3}}>24$

The first respondent asked if the OP was familiar with mathematical induction. The reply was that induction was the topic of the next chapter in her course.

Another suggested rationalizing the denominator to write:

$\displaystyle \frac{1}{2}\sum_{k=0}^{2499} \left(\sqrt{4k+3}-\sqrt{4k+1} \right)>24$

and then wished the OP good luck. She then asked for further help. She also asked for a proof by induction, which I provided as follows:

If I were going to use induction, I would state the hypothesis:

$\displaystyle \frac{1}{2}\sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{n+1}-1$

or equivalently:

$\displaystyle \sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{n+1}-2$ where $\displaystyle n\in\mathbb{N}_0$

base case $\displaystyle P_0$:

$\displaystyle \sum_{k=0}^0\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{0+1}-2$

$\displaystyle \sqrt{3}-\sqrt{1}>1-2$ true.

Consider:

$\displaystyle 6n+9=2\left(3n+\frac{9}{2} \right)$

$\displaystyle 6n+9=2\left(2(2n+3)-\frac{2n+3}{2} \right)$

$\displaystyle 6n+9=2\left(\sqrt{4(2n+3)^2}-\sqrt{\frac{(2n+3)^2}{4}} \right)$

$\displaystyle 6n+9>2\left(\sqrt{4(2n+3)^2-1}-\sqrt{\frac{(2n+3)^2+4n+3}{4}} \right)$

$\displaystyle 6n+9>2\left(\sqrt{(4n+5)(4n+7)}-\sqrt{(n+1)(n+3)} \right)$

$\displaystyle 8n+12-2\sqrt{(4n+5)(4n+7)}>2n+3-2\sqrt{(n+1)(n+3)}$

$\displaystyle (4n+7)-2\sqrt{(4n+5)(4n+7)}+(4n+5)>(n+2)-2\sqrt{(n+1)(n+3)}+(n+1)$

$\displaystyle (\sqrt{4n+7}-\sqrt{4n+5})^2>(\sqrt{n+2}-\sqrt{n+1})^2$

$\displaystyle \sqrt{4n+7}-\sqrt{4n+5}>\sqrt{n+2}-\sqrt{n+1}$

$\displaystyle \sqrt{4(n+1)+3}-\sqrt{4(n+1)+1}>\sqrt{(n+1)+1}-\sqrt{n+1}$

Now, adding this to the hypothesis, we have:

$\displaystyle \sum_{k=0}^n\left(\sqrt{4k+3}-\sqrt{4k+1} \right)+\sqrt{4(n+1)+3}-\sqrt{4(n+1)+1}>\sqrt{n+1}-2+\sqrt{(n+1)+1}-\sqrt{n+1}$

$\displaystyle \sum_{k=0}^{n+1}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\sqrt{(n+1)+1}-2$

$\displaystyle \frac{1}{2}\sum_{k=0}^{n+1}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{(n+1)+1}-1$

We have derived $\displaystyle P_{n+1}$ from $\displaystyle P_n$, thereby completing the proof by induction, and we may now state:

$\displaystyle \frac{1}{2}\sum_{k=0}^{2499}\left(\sqrt{4k+3}-\sqrt{4k+1} \right)>\frac{1}{2}\sqrt{2500}-1=24$

Another person replied with a technique using integration, which I am sure will be of little use to the OP.

I am just curious if there is a way to demonstrate the inequality is true by purely algebraic means. This is just for my own curiosity, and I will not post anyone's work there.