I am trying to show the numerical estimate $\Gamma(x) \leq 3x^x$ for $x \geq \frac{1}{2}$. I already know how to show $\Gamma(x) \leq x^x \leq 3x^x$ for $x \geq 1$. Therefore, it remains to show $\Gamma(x) = \int_0 ^\infty t^{x - 1} e^{-t} \,dt \leq 3x^x$ for $x \in [\frac{1}{2}, 1)$. My attempt is as follows, but it provides a constant that is greater than $3$:
Take $x \in [1/2, 1)$ and note $\Gamma(x) = \int_0 ^\infty t^{x - 1}e^{-t} \,dt = \int_0 ^1 t^{x - 1} e^{-t} \,dt + \int_1 ^\infty t^{x - 1} e^{-t} \,dt$. I will bound both integrals:
Note $$ \int_0 ^1 t^{x - 1} e^{-t} \,dt \leq \int_0 ^1 \frac{1}{t^{1/2}} \,dt = 2. $$ On the other hand, note that $$ \int_1 ^\infty t^{x - 1}e^{-t} \,dt \leq \int_1 ^\infty e^{-t} \,dt = \frac{1}{e}. $$ Knowing that $x^x \geq \frac{1}{\sqrt{2}}$, we have $Cx^x \geq \frac{C}{\sqrt{2}} \geq 2 + \frac{1}{e} \approx 2.4$. This means that $C$ needs to be chosen as approximately $\sqrt{2} \times 2.4 \approx 3.4 > 3$.
How can I fix this?
A rough estimate is $$ \Gamma(x) < 2 < 3x^x $$ for $1/2 \le x \le 1$.
The left inequality holds because $\Gamma(1/2) = \sqrt \pi$ and $\Gamma(1) = 1$ are both less than $2$, and $\ln \Gamma(x)$ is a convex function.
The right inequality holds because $x \mapsto x^x$ is increasing for $x \ge 1/e$, so that $$ 3x^x \ge 3 \left( \frac 12 \right)^{1/2} > 2 $$ for $x \ge 1/2$.