Prove that successive differences between square roots of integers decreases without using limits or derivatives

Question

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?

Answer 1

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.

Answer 2

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.