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I am trying to prove that $n \geq \sqrt{n+1} + \sqrt{n}$ for $n \geq 4$ (n in naturals of course). I am not sure if there are any specific inequalities that could help me out here. I also know that 4 is not exactly the "base" case, I just chose it arbitrarily as I am trying to compare two series. I attempted induction but am not sure if it is the correct way to do it.

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    $\begingroup$ The inequality in the title is different from the one in the question body. $\endgroup$
    – Martin R
    Feb 27, 2017 at 6:38

3 Answers 3

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Using that $4\leq n $, written as $2\leq\sqrt n $, we have $$ \sqrt {n-1}+\sqrt n\leq2\sqrt n\leq (\sqrt n)^2=n. $$

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Simply, $\sqrt {n-1}+\sqrt{n} \lt {2\sqrt n}$

$\forall n\in N$

For $n\ge 4,$ $2\sqrt n \le n$

Thus we have proved that $n \gt\sqrt {n-1}+\sqrt{n}$

$\forall n \ge 4$

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Since $a+b\leq\sqrt{2(a^2+b^2)}$, we obtain: $$\sqrt{n}+\sqrt{n-1}\leq\sqrt{2(n+n-1)}=\sqrt{4n-2}\leq\sqrt{4n}\leq n$$

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