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?
 A: What I would try first if I wanted it to be true - no tricks:
$n \ge \sqrt{n + 1} + \sqrt{n} \iff \frac{n}{\sqrt{n}} \ge \frac{\sqrt{n+1}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{n}} \iff \sqrt{n} \ge \sqrt{1 + \frac{1}{n}} + 1 \Leftarrow n \ge 5$.
A: To add to your step, observe the following:
$$\sqrt{n}+1 = \sqrt{(\sqrt{n}+1)^2} = \sqrt{n+1+2\sqrt{n}} > \sqrt{n+1+1} = \sqrt{n+2}.$$
The "$>$" part comes from your assumption $n \ge 5$, so $2\sqrt{n} \ge 2\sqrt{5} > 1$. Now, we have:
$$n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 = \sqrt{n+1} + (\sqrt{n}+1) > \sqrt{n+1} + \sqrt{n+2}.$$
A: 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.
A: Hint:  $\sqrt{n} + \sqrt{n+1} \leq 2\sqrt{n+1}$.  Can you take it from there?
A: Here is another way:
Define $f(x)=x-\sqrt{x}-\sqrt{x+1}$. We need to show that $f(x)\ge 0$ for all $x\ge 5$.
Since
$$f'(x)=1-\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x+1}} \ge 0, \quad \forall x\ge 1$$
the function is increasing on $[1,\infty)$. As $f(5)\ge 0$, the result follows.
A: You can replace the $\sqrt{n}$ on the RHS with another $\sqrt{n+1}$. Therefore you have $n\ge2\sqrt{n+1}$, or $n^2\ge4n+4$. $n^2-4n+4\ge8$, or $(n-2)^2\ge8$. The lowest integer solution to this is $5$, so $n\ge5.$
