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.

We start with
$$
\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}\binom{n}{k}
=\left\{\begin{array}{}
2^{n-1}&\text{if $n$ is odd}\\
2^{n-1}+\frac12\left(\raise{2pt}{n}\atop\frac{n}2\right)&\text{if $n$ is even}
\end{array}\right.\tag1
$$
Substitute $n\mapsto n-1$:
$$
\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k}
=\left\{\begin{array}{}
2^{n-2}&\text{if $n$ is even}\\
2^{n-2}+\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag2
$$
Subtract the $k=\left\lfloor\frac{n-1}2\right\rfloor$ term:
$$
\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor-1}\binom{n-1}{k}
=\left\{\begin{array}{}
2^{n-2}-\left(\raise{2pt}{n-1}\atop{\frac{n}2-1}\right)&\text{if $n$ is even}\\
2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag3
$$
Substitute $k\mapsto k-1$:
$$
\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1}
=\left\{\begin{array}{}
2^{n-2}-\frac12\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if $n$ is even}\\
2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag4
$$
Then
$$
\begin{align}
\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n}{k}k
&=n\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1}\tag{5a}\\
&=\left\{\begin{array}{}
n2^{n-2}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if $n$ is even}\\
n2^{n-2}-\frac{n}2\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag{5b}
\end{align}
$$
Explanation:
$\text{(5a)}$: $\binom{n}{k}k=\binom{n-1}{k-1}n$ and the $k=0$ term vanishes
$\text{(5b)}$: multiply $(4)$ by $n$

Double and add the middle term in the even $n$ case:
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}\min(k,n-k)
&=\left\{\begin{array}{}
n2^{n-1}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if $n$ is even}\\
n2^{n-1}-n\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if $n$ is odd}
\end{array}\right.\tag6
\end{align}
$$
The estimates for the central binomial coefficients from this answer give
$$
\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac13\right)}}\le\binom{n}{\frac{n}2}\le\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac14\right)}}\tag7
$$
which gives upper and lower bounds.
