$ 3 \sum^{n}_{k=1} \sqrt{k} \geq 2n \sqrt{n} + 1, n \in \mathbb{N} \setminus \{0\}$ Induction proof. How to separate multiplication. I have a problem with induction proofs like this one:
$$ 3 \sum^{n}_{k=1} \sqrt{k} \geq 2n \sqrt{n} + 1, n \in \mathbb{N} \setminus \{0\}$$
In the base case it works fine (for $n = 1$):
$$3 * \sqrt{1} \geq 2 + 1 \implies 3 = 3$$
Thesis ~ as above.
Now for $n + 1$:
$$ 3 \sum^{n+1}_{k=1} \sqrt{k} \geq 2(n+1) \sqrt{n+1} + 1$$
$$ 3 \sum^{n}_{k=1} \sqrt{k} + 3\sqrt{n+1} \geq 2n\sqrt{n+1} +1 + 2\sqrt{n+1}$$
And here is my problem: I don't know how to separate $1$ form $n$ in $2n\sqrt{n+1}$ on the left hand side.
On the right hand side of the inequality I got a sum which is good, but I can not separate that multiplication on the right. I thought about:

*

*using Bernoulis inequality, but then I "loose" the square root and can not use my induction thesis.

*changing $1$ under the square root for $1$ outside, but then the inequality doesn't work - right hand side is too big

*putting both sides of inequality in $e^{ln\{\}}$, but it doesn't work either

I don't know whot to do and it is not the first time I am having exactly the same problem in induction proofs. Can anybody help me?

For many problems of this kind, once you prove the base case you only need to compare the increment from $n$ to $n+1$ on both sides,
$$3\sqrt{n+1} \ge 2((n+1)\sqrt{n+1}-n\sqrt n) \\
\iff 2n\sqrt{n} \ge (2n-1)\sqrt{n+1}\\
\iff 4n^3 \ge (2n-1)^2(n+1)=4n^3-3n+1\\
\iff 3n\ge 1$$
which is true.
One added benefit is this can be easily converted to a non-induction, telescopic proof.
