Q:

Proving $2(2n-1)\ge n-1+\sqrt{(n-1)^2+4(n+1)^2}$

I am trying to prove that $2(2n-1)\ge n-1+\sqrt{(n-1)^2+4(n+1)^2}$ where $n\geq4$ is a Natural Number.
I plotted the graph for some values of $n$ and saw that it happens for $n\geq4$. But now I want to prove it more rigorously. This is what I have tried:
\begin{align*}
n-1+\sqrt{(n-1)^2+4(n+1)^2}&= n-1+\sqrt{5n^2+6n+5}\\
~&\leq n-1+\sqrt{5n^2+4n^2}\qquad(6n+5<4n^2\text{ for }n\ge3)\\
~&=n-1+\sqrt{9n^2}\\
~&=n-1+3n\\
~&=4n-1
\end{align*}
But I need it to be less that $4n-2$, so this is not how I should be proceeding. Can anyone give me some idea? Can it be done by induction?
Thank you.

A:

Instead of using induction, note what you're trying to prove can be simplified somewhat, such as follows
$$\begin{equation}\begin{aligned}
2(2n-1) & \ge n-1+\sqrt{(n-1)^2+4(n+1)^2} \\
4n - 2 & \ge n - 1 + \sqrt{(n^2 - 2n + 1) + (4n^2 + 8n + 4)} \\
3n - 1 & \ge \sqrt{5n^2 + 6n + 5}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Also, for $n \ge 4$,
$$\begin{equation}\begin{aligned}
(3n - 1)^2 - (5n^2 + 6n + 5) & = 9n^2 - 6n + 1 - (5n^2 + 6n + 5) \\
& = 4n^2 - 12n - 4 \\
& = 4(n^2 - 3n - 1) \\
& = 4(n(n-3) - 1) \\
& \ge 4(4(1) - 1) \\
& \gt 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus you have
$$(3n - 1)^2 \gt 5n^2 + 6n + 5 \implies 3n - 1 \gt \sqrt{5n^2 + 6n + 5} \tag{3}\label{eq3A}$$
This shows the final line in \eqref{eq1A} holds. Since you can reversibly get from that line back to the first line, it confirms your inequality holds.

