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

Question

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.

Answer 1

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.

Answer 2

So you have to prove $$3n-1\geq \sqrt{5n^2+6n+5}$$

Now square it (you can do that since both sides are positive): $$9n^2-6n+1\geq 5n^2+6n+5$$ so $$4n^2-12n-4\geq 0$$ or $$n^2-3n-1\geq 0$$ Since $f(x)=x^2-3x-1$ is increasing for $x\geq 3/2$ we have $$f(n) \geq f(4) = 16-12-1=3>0$$

Answer 3

we try to find inequality:

For any n>3, $n^2>3n+1$ ⇒ $2n^2-n>n^2+2n+1$

⇒$2n^2-n>(n+1)^2$ ⇒$\big(\frac{n}{n-1}\big)^2+\big(\frac{n}{n-1}\big)>\big(\frac{n+1}{n-1}\big)^2$

⇒ $1+4[\big(\frac{n}{n-1}\big)^2+\big(\frac{n}{n-1}\big)]>1+4\big(\frac{n+1}{n-1}\big)^2$

⇒$\big(1+\frac{2n}{n-1}\big)^2>1+\big(\frac{n+1}{n-1}\big)^2$

⇒ $1+\frac{2n}{n-1}>\sqrt{1+4\big(\frac{n+1}{n-1}\big)^2}$

⇒ $2+\frac{2n}{n-1}>1+\sqrt{1+4\big(\frac{n+1}{n-1}\big)^2}$

⇒$\frac{2(n-1+n)}{n-1}>1+\sqrt{\frac{(n-1)^2+4(n+1)^2}{(n-1)^2}}$

Multiplying both sides by $(n-1)$ we get:

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

The equality is only for $n=1$ ,so symbol $>$ is suitable for $ n≥4$.