The proof you've given is also the one I would have given (because it uses minimal resources), so I won't suggest an alternative - it just needs a little tweaking.
First, there's a minor typo: in two places, you've written $\sqrt{n + 1}$, where you meant $\sqrt{n - 1}$.
More seriously, you've assumed that $n$ is even: it doesn't make sense to talk of "$\frac{n}{2}$ terms" when $n$ is odd. This is easily remedied, because the idea of the proof is sound. One fix is as follows, although you may find a neater way:
Either $n = 2k - 1$ or $n = 2k$, where $k$ is a positive integer. The last $k$ terms of the sequence $1, \sqrt{2}, \ldots, \sqrt{n}$ are $\sqrt{n - k + 1}, \sqrt{n - k + 2}, \ldots, \sqrt{n}$. The smallest term is $\sqrt{n - k + 1}$, which is either $\sqrt{k}$ or $\sqrt{k + 1}$. Therefore, the sum of the $k$ terms is at at least $k\sqrt{k}$. Therefore it is at least $\frac{n}{2}\sqrt{\frac{n}{2}}$, because $k \geqslant \frac{n}{2}$.
Here's another simple idea, using only the very easily proved AM-GM inequality for two terms:
$$
\sum_{k=1}^n\sqrt{k} = \sum_{k=1}^n\frac{\sqrt{k} + \sqrt{n - k + 1}}{2} \geqslant \sum_{k=1}^n\sqrt[4]{k(n - k + 1)} \geqslant n^{5/4},
$$
because $k(n - k + 1) - n = (k - 1)(n - k) \geqslant 0$. Although very weak, it's enough for the job. (But I still prefer the solution given in the question.)
Looking for a proof using the Cauchy-Schwarz inequality, all I could come up with was this:
\begin{gather*}
\sqrt{k} - \sqrt{k - 1} = \frac{1}{\sqrt{k} + \sqrt{k - 1}} > \frac{1}{2\sqrt{k}} \quad (1 \leqslant k \leqslant n), \\
\therefore\ 1 + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}} < 2\sqrt{n}; \\
\text{but }
\left(1 + \sqrt{2} + \cdots + \sqrt{n}\right)
\left(1 + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}\right) \geqslant n^2,
\text{ by C-S}; \\
\therefore\ 1 + \sqrt{2} + \cdots + \sqrt{n} > \frac{n\sqrt{n}}{2}.
\end{gather*}
I came up with the following quite long but elementary argument as a
way of avoiding the use of Jensen's inequality.
For all $k \geqslant 1$, we have
$$
\sqrt{k + 1} - \sqrt{k} < \sqrt{k} - \sqrt{k - 1},
$$
because
$$
\left(\sqrt{k + 1} + \sqrt{k - 1}\right)^2 = 2k + 2\sqrt{k^2 - 1}
< 4k.
$$
Lemma
If $n > 2$, and
$a_2 - a_1 > a_3 - a_2 > \cdots > a_n - a_{n - 1} > 0$,
then
$$
\frac{a_1 + a_2 + \cdots + a_n}{n} > \frac{a_1 + a_n}{2}.
$$
Proof
Write $b_k = a_{k+1} - a_k$, so that $(b_k)_{1 \leqslant k < n}$ is
a strictly decreasing sequence.
For all $k$ such that $1 < k < n$, we have
\begin{gather*}
\frac{a_n - a_1}{a_k - a_1} =
\frac{b_1 + b_2 + \cdots + b_{n-1}}{b_1 + b_2 + \cdots + b_{k-1}}
= 1 + \frac{b_k + b_{k+1} + \cdots + b_{n-1}}
{b_1 + b_2 + \cdots + b_{k-1}} \\
\leqslant 1 + \frac{(n - k)b_k}{(k - 1)b_{k-1}}
< 1 + \frac{n - k}{k - 1} = \frac{n - 1}{k - 1}.
\end{gather*}
Setting $h = (a_n - a_1)/(n - 1)$ and $a_k' = a_1 + (k - 1)h$, we
therefore have $a_1 = a_1'$, $a_n = a_n'$, and $a_k > a_k'$
($1 < k < n$), whence
$$
a_1 + a_2 + \cdots + a_n > a_1' + a_2' + \cdots + a_n' =
n\frac{a_1' + a_n'}{2} = n\frac{a_1 + a_n}{2},
$$
as required.
$\square$
Taking $a_k = \sqrt{k}$ in the lemma, we get
$$
1 + \sqrt{2} + \cdots + \sqrt{n} >
\frac{n\left(\sqrt{n} + 1\right)}{2} \quad (n > 2).
$$
(Even after all this, I still prefer the OP's own proof - but
looking for alternatives is fun.)
Here is another quite simple proof, which I think I prefer to
the one given in the question. (Looking at the other answers, I see that it is a variant of a proof that Barry Cipra has already posted, but there are probably enough differences to make it still worth posting.)
If $n \geqslant 4$, there is a unique positive integer $r$ such that
$(r + 1)^2 \leqslant n < (r + 2)^2$, and:
\begin{align*}
& \phantom{=} 1 + \sqrt{2} + \cdots + \sqrt{n}
\\ & \geqslant 1 + (\sqrt{2} + \sqrt{3} + \sqrt{4}) + \cdots
+ (\sqrt{r^2 + 1} + \cdots + \sqrt{(r + 1)^2})
\\ & > 0 + (2^2 - 1^2) \cdot 1 + (3^2 - 2^2) \cdot 2 + \cdots
+ ((r + 1)^2 - r^2) \cdot r
\\ & > 2 \cdot 1^2 + 2 \cdot 2^2 + \cdots + 2 \cdot r^2
\\ & = \frac{1}{3}r(r + 1)(2r + 1) > \frac{2}{3}r^3
\\ & > \frac{2}{3}\left(\sqrt{n} - 2\right)^3.
\end{align*}
Although there's no need (for present purposes), one could also
argue that for all $n$, there is a unique positive integer $s$ such
that $(s - 1)^2 < n \leqslant s^2$, and:
\begin{align*}
& \phantom{=} 1 + \sqrt{2} + \cdots + \sqrt{n}
\\ & \leqslant 1 + (\sqrt{2} + \sqrt{3} + \sqrt{4}) + \cdots
+ (\sqrt{(s - 1)^2 + 1} + \cdots + \sqrt{s^2})
\\ & \leqslant (1^2 - 0^2) \cdot 1 + (2^2 - 1^2) \cdot 2
+ (3^2 - 2^2) \cdot 3 + \cdots + (s^2 - (s - 1)^2) \cdot s
\\ & < 2 \cdot 1^2 + 2 \cdot 2^2 + 2 \cdot 3^2 + \cdots + 2 \cdot s^2
\\ & = \frac{1}{3}s(s + 1)(2s + 1) < \frac{2}{3}(s + 1)^3
\\ & < \frac{2}{3}\left(\sqrt{n} + 2\right)^3.
\end{align*}
