Prove by Induction: $n \le 3 \sqrt{n} +4$. How to work with the Square-root?

Question

I want to prove the statement $$n \le 3 \sqrt{n} +4$$

for every $n$ belongs to $N$ by induction.

So what I have done so far is proving for $p(1)$ is true and assuming that $p(n)$ is true. Now, I want to prove that $p(n+1)$ is also true; $$n+1 \le 3 \sqrt{n+1} +4 +1$$

Well basically I add $1$ to both side of the inequation to get $n+1$ on the left hand side, but I am not sure how to work with the square-root. Should I leave the $n$ under the square-root unchanged in this step? and what should I do next?

Answer 1

It is false. Take $n = 100$. Then: $$100 = n > 3\sqrt{n} + 4 = 34.$$

Answer 2

With $x^2=n$ (which is valid for $n\geq 0$), you have $$ n-3\sqrt{n}-4=x^2-3x-4=(x+1)(x-4)\geq 0 $$ whenever $x\geq 4$. So in particular, when $n>16$, your inequality is false.