Is it true that $ n\sqrt{n^2-1} > \frac{n^2}{2}?$

Question

Is it true?

$$ n\sqrt{n^2-1} > \frac{n^2}{2}$$

Even if $ n \rightarrow \infty$?

Answer 1

Hint. One may use that $$ \sqrt{1-\frac1{n^2}}> \frac12,\qquad n\ge2, $$ giving, by multiplying by $n^2$, $$ n^2\sqrt{1-\frac1{n^2}}=n\sqrt{n^2-1}> \frac{n^2}2,\qquad n\ge2. $$

Answer 2

If $n > 0$, if $\sqrt{n^2-1} \in \mathbb{R}$ we must have $n\ge 1$. Assuming that, we divide both sides by $n$ to get $$ \sqrt{n^2-1} > n/2 $$ which is equivalent to (since $n>1$) $$ n^2-1 > n^2/4 $$ which is equivalent to $$n > \frac{2\sqrt{3}}{3}\approx 1.1547.$$

We conclude that as long as $n>1.1547$, your inequality holds.

Answer 3

$$n\sqrt{n^2-1}=n^2\sqrt{1-\frac{1}{n^2}}$$ Using series expression at '$n=\infty$' we get that $$n^2\sqrt{1-\frac{1}{n^2}}=n^2\left(1-\frac{1}{2n^2}-\frac{1}{8n^4}+O\left(\frac{1}{n^5}\right)\right)$$ $$=n^2-\frac{1}{2}+O\left(\frac{1}{n^2}\right)$$ So it's bigger than $\frac{n^2}{2}$ for large enough $n$.

Answer 4

As shown above by "gt6989b", $n>1.15$

Suppose $n>1.8, n^2>2$ Then $n^2-2>0$

$n^2-2>0$

$(n^2-1)-1>0$

$3(n^2-1)-1>0$

$3n^2-3-1>0$

$3n^2-4>0$

$3n^2+n^2-4>n^2$

$4n^2-4>n^2$

$n^2-1>\frac{n^2}{4}$

$\sqrt{n^2-1}>\sqrt{\frac{n^2}{4}}$

$\sqrt{n^2-1}>\frac{n}{2}$ or $n\sqrt{n^2-1}>\frac{n^2}{2}$

Answer 5

I'm not sure what "even if", $n \to \infty$. I'd say "especially if" $n \to \infty$.

If $n>2$ then $n^2 > 2n > 2$ and $n\sqrt{n^2 - 1} = $

$n\sqrt{n^2 - 2+1} > n\sqrt{n^2 - 2n + 1} = $

$n\sqrt{(n-1)^2}= n(n-1)=$

$n^2 - n > n^2 - \frac {n^2}2 = \frac {n^2}2$

And that's it....

It is true if $n> 2$. Period.

I suppose you're thinking is that if $n\to \infty$ is somehow equivalent to $n = \infty$ and if $n$ is infinity than both expressions are infinite and we can't have one infinite value greater than another value.

But that not what $n\to \infty$ means. Infinity is not a number that we can plug into $n$. $n \to \infty$ simply means that $n$ may be as arbitrarily large as we like.

If "$n$ approaches $\infty$" that means we may assume $n > 2$ so this it true. Notice if $1\le n <\frac {2\sqrt{3}}3$ or $n\le -1$ then $n\sqrt{n^2 - 1} < \frac {n^2}2$