Lillianna Andersen

2022-07-08

Inductive Proof that $\sqrt{k+1}-\sqrt{k}\le \frac{1}{\sqrt{k+1}}$
Help me understand this:
$\sqrt{k+1}-\sqrt{k}\le \frac{1}{\sqrt{k+1}}$

Dalton Lester

Expert

Step 1
For $K\ge 0$ we have ${K}^{2}\le {K}^{2}+K=K\left(K+1\right)$ and so $K\le \sqrt{K\left(K+1\right)}$ which further implies $K+1\le \sqrt{K\left(K+1\right)}+1,$.
Step 2
Thus we have $K+1-\sqrt{K\left(K+1\right)}\le 1,$, or $\sqrt{K+1}\left(\sqrt{K+1}-\sqrt{K}\right)\le 1,$, which is the required inequality.

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