Here is a proof.
First evaluate the integral. I use the generating function for squares of Laguerre polynomials ($I_0$ the modified Bessel function):
$$
\sum_{n=0}^{\infty}L_{n}(x)^2 z^n = \frac{1}{1-z} \exp\left(-\frac{2 x z}{1-z}\right) I_{0}\left(\frac{2 x \sqrt{z}}{1-z}\right)
$$
for $|z|<1$ (All the formulas I use, one can find it in e.g. Gradshteyn and Ryzhik.)
This gives for the integral
$$
T_{n} = \int_{0}^{\infty} dx \sqrt{x} e^{-x} L_{n}(x)^2 = [z^n] \frac{1}{1-z} \int_{0}^{\infty} dx \sqrt{x} \exp\left(-x \frac{1+z}{1-z}\right) I_{0}\left(x \frac{2 \sqrt{z}}{1-z}\right)
$$
$[z^n]$ means, as usual, "take the coefficient of $z^n$ in the (formal) power expansion in $z$ of the following expression".
The integral can be evaluated (it is in Gradshteyn and Ryzhik) and the result is
$$
T_{n} = [z^n] \frac{\sqrt{\pi}}{2} (1-z)^{-1} P_{1/2}\left(\frac{1+z}{1-z}\right)
$$
with $P_{1/2}(x)$ the Legendre Function.
Expanding the Legendre function and the prefactor in powers of $z$ gives for the coefficient of $[z^n]$
$$
T_{n} = \frac{1}{8\sqrt{\pi}}\sum_{m=0}^{n}\frac{\Gamma\left(m-\frac{1}{2}\right) (2(n-m)+1)!}{4^{n-m} m!^2 (n-m)!^2}
$$
Now I was lazy and asked Mathematica to simplify, which gives
$$
T_{n}= \frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} {_3}F_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right)
$$
with the generalized hypergeometric function ${_3}F_{2}$ evaluated at $1$.
With the exact solution at hands we are almost done. We observe that the sum defining the hypergeometric function
$$
{_3}F_{2}\left(-\frac{1}{2},-\frac{1}{2},-n;1,-\frac{1}{2}-n;1\right)
=\sum_{m=0}^{\infty} \frac{\left(-\frac{1}{2}\right)_{m}^{2}(-n)_m}{m!^2 \left(-n-\frac{1}{2}\right)_{m} }
$$
has only positive terms (the minus signs in the Pochhammer symbols $(-n)_m$ and $(-n-\frac{1}{2})_m$ cancel.). We get a lower bound by taking only the first two expansion terms of ${_3}F_{2}$ and the well known estimate for the Gamma functions in front
$$
\frac{\Gamma(n + \frac{3}{2})}{\Gamma(n+1)} \geq n^{1/2}
$$
and find
$$
T_{n} \geq n^{1/2} \left(1+\frac{n}{2(n+1)}\right)
$$
which is larger than $\sqrt{n+1}$ for $n\geq 2$.
Btw, the asymptotic ($n\rightarrow \infty$) value of $T_n$ is
$$
T_{n}\sim \frac{4}{\pi} n^{1/2}
$$
I am pretty sure that the calculation can be done more elegantly, but had not the time to dig deeper.
Edit:
With the same trick one finds more generally:
$$
\int_{0}^{\infty} dx \, x^{\nu} e^{-x} L_{n}(x)^2 = \frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {_3}F_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {_2}F_{1}(-\nu,-\nu;1;1)) (1+ O(n^{-1}))
$$
The last term simplifies to:
$$
\frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} \frac{\Gamma(2 \nu +1)}{\Gamma(\nu+1)^2} (1+ O(n^{-1}))
$$
for $\nu > 0$.
Edit:
After some massaging with Mathematica I could determine the coefficient of $n^{-1}$ in the above expansion:
$$
\frac{\Gamma(n+\nu+1)}{\Gamma(n+1)} {_3}F_{2}\left(-n,-\nu,-\nu;1,-n-\nu;1\right) = n^{\nu} {_2}F_{1}(-\nu,-\nu;1;1)) (1 + \frac{\nu}{2 n} + O(n^{-1-2 \nu}))
$$
The determination of the next order coefficient as well as the exact exponent is rather cumbersome. One could do better than $O(n^{-1-2 \nu})$.
