Show that $n \ge \sqrt{n+1}+\sqrt{n}$
Solution 1:
Hint: $\sqrt{n} + \sqrt{n+1} \leq 2\sqrt{n+1}$. Can you take it from there?
Solution 2:
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.