Prove $\sqrt{n!} \gt \frac{n}{2}$ $\forall$ $n \ge 1$

Question

Prove that $$\sqrt{n!} \gt \frac{n}{2}$$ $\forall$ $n \ge 1$

My try:

I Tried using Induction:

$P(1)$ is True obviously.

Let $P(k)$ Be True Then we have

$$\sqrt{k!} \gt \frac{k}{2}$$

Now $$\sqrt{(k+1)!}=\sqrt{k+1}\sqrt{k!}\gt \sqrt{k+1}\frac{k}{2}$$

Now Since $x^2-x-1 \gt 0$ $\forall $ $x \ge 2$ We have

$$k^2 \gt k+1$$ $\forall$ $k \ge 2$

Hence

$$k \gt \sqrt{k+1}$$

Hence

$$\sqrt{(k+1)!} \gt \sqrt{k+1}\frac{k}{2}\gt \frac{1}{2}\sqrt{k+1}\sqrt{k+1}=\frac{k+1}{2}$$

Hence Proved

Is there any alternate approach?

Answer 1

For all $n>1$ it is enough to prove that $n(n-1) > n^2/4$. Iff $3 n^2/4 > n$. For $n=2$ it is true. $n^2$ grows faster than $n$ so it is also true for larger $n$.

For $n=1$ we can see it is true by substitution.

Answer 2

Assume $n \geq 2$, Then, $$\frac{4}{3} < n$$ $$\Leftrightarrow n < 4(n-1)$$ $$\Leftrightarrow \frac{n}{4} < (n-1).$$ Then $$\frac{n}{4} < (n-1) \leq (n-1)!$$ Multiply by $n$ and take the square root $$ \frac{n}{2} < \sqrt{n!}$$