How to prove that $\operatorname{lcm}\{1,\ldots,n\}\geq (\sqrt{n})^{\pi(n)}$? Let $\operatorname{LCM}[n]:=\operatorname{lcm}\{1,\ldots,n\}$. It is easy to verify $\operatorname{LCM}[n]\geq 2^{\pi(n)}$, where $\pi(n)$ counts the number of distinct primes up to $n$. 
But how can I prove the bound $\operatorname{LCM}[n]\geq (\sqrt{n})^{\pi(n)}$?
For the bound $2^{\pi(n)}$, I used the identity:
$$\operatorname{LCM}[n]=\!\!\!\prod_{p\in\mathbb{P},p\leq n} p ^{\lfloor \log_p(n) \rfloor}$$
But this doesn't work with the lower bound of $\operatorname{LCM}[n]\geq (\sqrt{n})^{\pi(n)}$.
Can someone give me a hint?
 A: The exact same identity works.  You can do much better than $p^{\lfloor \log_p(n) \rfloor} \ge 2$, and show that $p^{\lfloor \log_p(n) \rfloor} > \sqrt{n}$.  Here's the idea: $p^{\lfloor \log_p(n) \rfloor}$ is just the largest power of $p$ that fits into $[1,n]$.
If $p$ is in the range $(n^{1/2},n]$ there is nothing to prove.  But if $p$ is in $(n^{1/3},n^{1/2}]$ then $p^2 \le n$ and $p^2 > \sqrt{n}$.  If you think about it, no matter what prime you start with, there is some power of $p$ that is larger than $\sqrt{n}$.
A: Not sure why I can't comment here, but anyway... Here's a suggestion:
From your bound, you get $LCM[n]\ge\prod_{p\in\mathbb P, p\le n}p$.
It would therefore suffice to show that the geometric mean of $\{p\in\mathbb P,p\le n\}$ is greater than $\sqrt n$. Pairing the primes, the smallest with the biggest etc, the product of each pair is greater than $n$ (by Bertrand's postulate) so you're done?
A: All that is needed for the proof is the elementary inequality $$\lfloor x\rfloor \geq \dfrac{x}{2} \text{   for $x \geq 1$}$$
Note that the $\text{lcm}(\{1,2,\ldots,n\}) = \exp(\psi(n))$, where $\psi(n)$ is the second Chebyshev function. Hence, we want to prove that $$\exp(\psi(n)) \geq \exp \left( \log(\sqrt{n}) \pi(n)\right) = \exp \left( \dfrac{\pi(n) \log(n)}2 \right)$$
So, we need to prove that $\psi(n) \geq \dfrac{\pi(n) \log(n)}2$.
Note that $$\psi(n) = \sum_{p\text{ is prime}} \sum_{\overset{k \in \mathbb{Z}^+}{p^k \leq n}} \log_e(p) = \sum_{p\text{ is prime}} \log_e(p) \lfloor \log_p(n)\rfloor$$
Now for $x\geq 1$, we have $\lfloor x\rfloor \geq \dfrac{x}{2}$. Since $p \leq n$, we have $\lfloor \log_p(n)\rfloor \geq \dfrac{\log_p(n)}{2}$.
Hence, we get that
$$\psi(n) = \sum_{p\text{ is prime}} \log_e(p) \lfloor \log_p(n)\rfloor \geq \sum_{p\text{ is prime}} \dfrac{\log_e(p) \log_p(n)}{2} = \sum_{p\text{ is prime}} \dfrac{\log_e(n)}{2} = \dfrac{\pi(n)\log_2(n)}2$$
A: This is a slightly different, and hopefully simpler, take on this answer.
We have that for $x\ge1$, $\lfloor x\rfloor\gt\frac x2$. Thus, for $n\gt1$, $n$ has at least one prime factor, so
$$
\begin{align}
\mathrm{lcm}(1,2,3,\dots,n)
&=\prod_{\substack{p\le n\\p\text{ prime}}}p^{\left\lfloor\log(n)/\log(p)\right\rfloor}\\
&\gt\prod_{\substack{p\le n\\p\text{ prime}}}p^{\frac12\log(n)/\log(p)}\\
&=\prod_{\substack{p\le n\\p\text{ prime}}}n^{1/2}\\
&=\left(n^{1/2}\right)^{\pi(n)}
\end{align}
$$
