Show that $n \ge \sqrt{n+1}+\sqrt{n}$ (how) Can I show that:
$n \ge \sqrt{n+1}+\sqrt{n}$ ?
It should be true for all $n \ge 5$.
Tried it via induction:


*

*$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true. 

*$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$
Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)
Is this the right way?

First observe that since $5 \le n$, we have:
$$
4(n+1) =4n+4 < 4n+5 \le 4n+n = 5n \le (n)n = n^2
$$
Hence, since $4(n+1)<n^2 \Rightarrow \boxed{n+1<\dfrac{1}{4}n^2}$ (and $f(x)=\sqrt{x}$ is monotonically increasing and $n\ge0$), we have:
$$
\sqrt{n+1}+\sqrt{n} \le \sqrt{n+1}+\sqrt{n+1} = 2\sqrt{n+1}<2\sqrt{\dfrac{1}{4}n^2}=2\left(\dfrac{1}{2}n\right) = n
$$
as desired.
