Proving $\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$ I would like to know how to prove the following assertion :

For every $n>0$: $$\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$$

 A: There are many approaches. One of them is to try to prove the equivalent assertion that $\sqrt{n+1}-\sqrt{n} \le \frac{1}{\sqrt{n+1}}$.
Note that 
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}.\tag{$1$}$$
It is clear that for $n \gt 0$, the right-hand side of $(1)$ is $\lt \frac{1}{\sqrt{n+1}}$.
The calculation gives a stronger estimate. The right-hand side of $(1)$ is actually close to $\frac{1}{2\sqrt{n+1}}$ when $n$ is large.   
Another way: Equivalently, we want to prove that $\sqrt{n+1}-\frac{1}{\sqrt{n+1}} \le \sqrt{n}$. 
We have 
$$\sqrt{n+1}-\frac{1}{\sqrt{n+1}}=\frac{(n+1)-1}{\sqrt{n+1}}=\frac{n}{\sqrt{n+1}}.$$
Since $\sqrt{n+1} \gt \sqrt{n}$, we conclude that if $n \gt 0$ then $\frac{n}{\sqrt{n+1}}\lt \sqrt{n}$.  
A: $$\sqrt{n}+\frac{1}{\sqrt{n+1}}\geq \sqrt{n+1}$$
$$\sqrt{n}\sqrt{n+1}+1\geq n+1$$
$$\sqrt{n}\sqrt{n+1}\geq n$$
$$n(n+1)\geq n^2$$
$$n^2+n\geq n^2$$
$$n\geq 0$$
A: Or you could obfuscate the problem. Let $t=\sqrt{n+1}$. Then we have $t>1$ and $\sqrt{n}=\sqrt{t^2-1}$, so
$$
\begin{align*}
 &\sqrt{t^2-1}+\frac{1}{t} \ge t\\
\text{iff}\qquad &\sqrt{t^2-1}\ge t-\frac{1}{t}\\
\text{iff}\qquad &t^2-1 \ge t^2-2+\frac{1}{t^2}\\
\text{iff}\qquad &1 \ge \frac{1}{t^2}\\
\text{iff}\qquad &t^2\ \ge 1
\end{align*}
$$
which is true, since $t>1$.
