Proving $\frac{1}{2}(n+1)<\frac{n^{n+\frac{1}{2}}}{e^{n-1}}$ without induction I want to show that $\displaystyle\frac{1}{2}(n+1)&lt;\frac{n^{n+\frac{1}{2}}}{e^{n-1}}$. But except induction, I do not know how I could prove this?
If $n\ge3$, $\frac12(n+1)&lt;n$ and $n\ge\mathrm{e}$ hence $n^{n-1}\ge\mathrm{e}^{n-1}$ and it is enough to prove that $n&lt;n^{3/2}$, which is obviously true. Then you can check manually the cases $n=1$ (for which the strict inequality is false, by the way) and $n=2$.
This is equivalent to showing $$n \leq 2 \frac{n^{n+\frac{1}{2}}}{e^{n-1}}-1 =  2\left(\frac{n^{n}n^{1/2}e}{e^n}\right)-1 $$ 

Let $$f(n) = 2\left(\frac{n^{n}n^{1/2}e}{e^n}\right)-n-1$$

Then $f(n) \geq 0$ (i.e. find critical points and use first derivative test).