Showing $\Gamma\left(x+1\right) \ge x^{x/2}$, when $x\ge2$ It is easy to show that $(2n)! \ge (2n)^n$.
Indeed, if we have $ 1\le k\le n$, for integer $k,n$ , we can write:
$$
\begin{aligned}
0\lt n+\frac{1}{2}-k &\le n - \frac{1}{2} \\
\left(n+\frac{1}{2}-k\right)^2 &\le \left(n - \frac{1}{2}\right)^2 \\
k\cdot(2n-k+1)&\ge2n
\end{aligned}
$$
and so
$$
(2n)! = 1\cdot2\cdots (2n) = \prod_{k=1}^{n} {k\cdot(2n-k+1)} \ge (2n)^n
$$
A variation of the above argument can be used to prove the statement for odd integers.
Is this statment true for all real $x\ge2$? 
 A: Let
$$f(x) =  \log \Gamma(x + 1) - \frac{x \log(x)}{2} = \log(x) + \log \Gamma(x) - \frac{x \log(x)}{2}$$
We will prove that $f$ is a convex function with $f(2) = 0$ and $f'(2) > 0$. Using the subderivative property of convex functions we then get 
$$f(x) \geq f(2) + (x - 2) f'(2) \ge 0 \qquad \text{for } x \ge 2,$$
which is what we wanted to prove.
The first two derivatives of $f$ are
$$\begin{align*}
f'(x) &= \psi(x) + \frac{1}{x} - \frac{1}{2} - \frac{\log(x)}{2} \\
f''(x) &= \psi'(x) - \frac{1}{x^2} - \frac{1}{2x}
\end{align*},$$
where $\psi$ resp. $\psi'$ are the di- resp. trigamma function. We will now use the formulae from Abramowitz and Stegun.
First of all, formula 6.3.2 tells us
$$f'(2) = \psi(2) - \frac{\log(2)}{2} = 1 - \gamma - \frac{\log(2)}{2} = 0.076... > 0$$
To prove convexity, we will show $f''(x) \ge 0$ for all $x \ge 2$. From 6.4.10 we get the representation
$$\psi'(x) = \sum \limits_{k = 0}^\infty \frac{1}{(x + k)^2} = \frac{1}{x^2} + \sum \limits_{k = 1}^\infty \frac{1}{(x + k)^2}.$$
Now it only remains to prove the inequality
$$\sum \limits_{k = 1}^\infty \frac{1}{(x + k)^2} \ge \frac{1}{2x}.$$
As pointed out by Jack Lam in the comments (many thanks!) this can be proven using telescoping:
$$\sum \limits_{k = 1}^\infty \frac{1}{(x + k)^2} \ge \sum \limits_{k = 1}^\infty \frac{1}{(x + k)(x + k + 1)} = \sum \limits_{k = 1}^\infty \left(\frac{1}{x + k} - \frac{1}{x + k + 1}\right) =  \frac{1}{x + 1} \ge \frac{1}{2x}.$$
