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I have been trying to prove that $$\sqrt{n+1} - \sqrt{n} > \sqrt{n+2} - \sqrt{n+1}$$

for ALL $n = 1,2,3,...$ without success.

A similar question has been asked here Why does the difference in square roots of two consecutive integers gets smaller as n grows?. However, the answers are either intuitive i.e. not rigorous, or use limits/derivatives.

My problem with using limits is that it only tells you this will eventually be true. Can we show rigorously that this is true for all $n = 1,2,3,...$ by just using algebra and basic proof methods?

user35687
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  • Did you see AndrewLi's comment beneath that post? It gives the best explanation imo. – MyMolecules Oct 18 '21 at 16:33
  • @MyMolecules I did. It's a nice explanation for intuition but it's not a proof. – user35687 Oct 18 '21 at 16:34
  • Did you check all answers to the other question? There are at least two which do not use limits, derivatives, or intuition. – And if you think that an answer is unclear: you have sufficient reputation to leave a comment and ask for clarification. – Martin R Oct 18 '21 at 17:00

2 Answers2

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Just square that!
$$(\sqrt{n+2}+\sqrt n)^2 = 2n+2+2\sqrt{n^2+2n} < 2n+2+2\sqrt{n^2+2n+1} = 4n+4 = (2\sqrt{n+1})^2$$ $$\sqrt{n+2}+\sqrt n < 2\sqrt{n+1}$$ Some steps are left, but they are obvious.

MH.Lee
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The answers in the link are sufficient, since the identity

$$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}}$$ immediately implies $$\sqrt{n+2} - \sqrt{n+1} = \frac{1}{\sqrt{n+2} + \sqrt{n+1}},$$ and since $$\sqrt{n+2} > \sqrt{n},$$ it follows that $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} > \frac{1}{\sqrt{n+2} + \sqrt{n+1}} = \sqrt{n+2} - \sqrt{n+1}.$$

I don't know why you think this is not rigorous.

heropup
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