Inequality involving sums with binomial coefficient I am trying to show upper- and lower-bounds on
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)$$
(where $n\geq 1$) in order to show that it basically grows as $\Theta(n)$.
The upper-bound is easy to get since $\min(i, n-i)\leq i$ for $i\in\{0, \dots n\}$ so that
$$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)\leq \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}i = \frac{n}{2}.$$
Thanks to Desmos, I managed to find a lower bound, but I am struggling to actually prove it. Indeed, I can see that the function $f(n)=\frac{n-1}{3}$ does provide a lower-bound. One can in fact rewrite
$$\frac{n-1}{3}=\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\frac{2i-1}{3}.$$
I was thus hoping to show that for each term we have $\frac{2i-1}{3}\leq \min(i, n-i)$, but this is only true if $i\leq \frac{3n+1}{5}$ and not generally for $i\leq n$. I imagine there is a clever trick to use at some point but for some reason I am stuck here.
Any help would be appreciated, thank you!
EDIT: Thank you everyone for all the great and diverse answers! I flagged River Li's answer as the "accepted" one because of its simplicity due to the use of Cauchy-Schwartz inequality, which does not require a further use of Stirling's approximation. Note that the other answers which involve such an approximation are much tighter though, but proving $\Theta(n)$ growth was sufficient here.

Let's first note that $\binom{n}{i}\cdot i = n\cdot \binom{n-1}{i-1}$.
For odd $n=2m+1$, this makes
$$
\begin{align}
S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i)
&= 2\sum_{i=0}^m\binom{n}{i}\cdot i
 = 2n\sum_{i=1}^m\binom{n-1}{i-1}
 = 2n\sum_{j=0}^{m-1}\binom{2m}{j}
\\
&= n\cdot\left( \sum_{j=0}^{2m}\binom{2m}{j} - \binom{2m}{m} \right)
 = n\cdot\left( 2^{2m} - \binom{2m}{m} \right)
\end{align}
$$
where we have used that $\binom{2m}{j}=\binom{2m}{2m-j}$. This makes
$$
\frac{S_n}{2^n}
= \frac{n}{2}\cdot\left (1 - \frac{\binom{2m}{m}}{2^{2m}} \right)
\approx \frac{n}{2}\cdot\left( 1 - \frac{1}{\sqrt{\pi m}} \right).
$$
For even $n=2m$, we get
$$
\begin{align}
S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i)
&= 2\sum_{i=0}^m\binom{n}{i}\cdot i - \binom{2m}{m}\cdot m
 = 2n\sum_{i=1}^m\binom{n-1}{i-1} - \binom{2m}{m}\cdot m
\\
&= n\sum_{j=0}^{n-1}\binom{n-1}{j} - \binom{2m}{m}\cdot m
 = n\cdot\left( 2^{n-1} - \frac{1}{2}\binom{2m}{m} \right)
\end{align}
$$
which once more makes
$$
\frac{S_n}{2^n}
= \frac{n}{2}\cdot\left(1-\frac{\binom{2m}{m}}{2^{2m}}\right).
$$
In both cases, you get $n/2$ as an upper bound. However, there are strong bounds on $\binom{2m}{m}/2^{2m}$ which can be applied:
$$
\frac{e^{-1/8m}}{\sqrt{\pi m}}
\le
\frac{\binom{2m}{m}}{2^{2m}}
\le \frac{1}{\sqrt{\pi m}}
$$
Eg, see Jack D'Aurizio's derivation of this, or Wikipedia.

Additional bounds have been provided by robjohn. The following bounds seem to be the tightest proven so far:
$$
\frac{4^me^{-1/8m}}{\sqrt{\pi m}}
< \binom{2m}{m}
< \frac{4^m}{\sqrt{\pi\left(
   m+\frac{1}{4}
  \right)}}
$$
The following bound is even tighter, but I have no proof of it, just numerical evidence:
$$
\frac{4^m}{\sqrt{\pi\left(
   m+\frac{1}{4}+\frac{1}{32m}
  \right)}}
< \binom{2m}{m}
$$
It's the same as the above up to second order approximation, so not a bit difference, but easier to compute.

