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I am trying to prove an inequality, but that itself is not the question here. Looking around for ideas how to proceed I found solutions for the same problem. But in several occasions I couldn't follow the answers because of this step. In both of these excerpts (taken from answered questions on this site, the links are below) two different square roots are combined like this. \begin{align} \sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}\tag{Now it's an inequality} \end{align}

Why is this possible even though the square roots are not identical?

Here the excerpts:

\begin{align} = & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}\right] \quad \text{multiply conjugate}\\ > & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{2\sqrt{n + 1}}\right] \\ \end{align} taken from here

or

\begin{align} \sqrt{k+1}-\sqrt{k}&=\frac{1}{\sqrt{k+1}+\sqrt{k}}\\ &\le\frac{1}{2\sqrt{k}} \end{align}

taken from here

I hope someone can explain this to me or point me in the right direction

amWhy
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cyanit42
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    I haven't read the body of your question yet, but the title is horribly wrong. $\sqrt{n}+\sqrt{n+1}$ is not equal to $2\sqrt{n}$ though it is going to be larger than $2\sqrt{n}$ – JMoravitz Nov 05 '20 at 19:28
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    @beta314 Are you aware of the meaning of the symbols $>$ and $\le$? Because you seem to assume that they mean "$=$". –  Nov 05 '20 at 19:28
  • Because $n<n+1$ and $x\mapsto\sqrt{x}$ is strictly increasing. – Michael Hoppe Nov 05 '20 at 19:41

2 Answers2

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No, the linked examples use $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, which is equivalent to the trivial $\sqrt{n+1}>\sqrt{n}$.

J.G.
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We have that

$$\sqrt{n} + \sqrt{n + 1} >\sqrt{n}+\sqrt{n}= 2\sqrt{n}$$

since $\sqrt{n + 1}>\sqrt{n}$ or also dividing by $\sqrt n$

$$\sqrt{n} + \sqrt{n + 1}> 2\sqrt{n} \iff 1+\sqrt{1+\frac1n}>2$$

which is true since $\sqrt{1+\frac1n}>1$.

user
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