<s>[INST] Why is $\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}$? 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
 [/INST] No, the linked examples use $\sqrt{n}+\sqrt{n+1}>2\sqrt{n}$, which is equivalent to the trivial $\sqrt{n+1}>\sqrt{n}$.
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