Prove that for any $a\geq \frac 12$, $$ \int_0^{a} x^{a-1}e^{-x}dx > \frac 12 \Gamma(a)$$
Using the incomplete gamma function, this rewrites as $$ \frac{\gamma(a,a)}{\Gamma(a)}>\frac 12$$
This inequality appears when I tried to show a property of the $\chi^2$ distribution. I verified that it is true numerically. It also seems that the lower bound is tight: $$\lim_{a\to \infty}\frac{\gamma(a,a)}{\Gamma(a)}=\frac 12$$
I don't have much experience working with special functions and I've made no significant progress towards a proof.
Substituting $x=a(1+y)$ in the defining integrals, we have $$\frac{\gamma(a,a)}{\Gamma(a)}=\frac{\int_{-1}^0(1+y)^{a-1}e^{-ay}\,dy}{\int_{-1}^\infty(1+y)^{a-1}e^{-ay}\,dy}.$$ Now let $y-\log(1+y)=z^2/2$ so that $z\gtrless 0$ when $y\gtrless 0$. Then $y\mapsto z(y)$ is a bijection, $\color{blue}{z(y)<y}$ for all $y\neq 0$, and $\big(1-1/(1+y)\big)\,dy=y\,dy/(1+y)=z\,dz$, so that, writing $\int_{-1}^\infty=\int_{-1}^0+\int_0^\infty$, we get $$\frac{\gamma(a,a)}{\Gamma(a)}=\frac{1}{1+f(a)},\quad f(a)=\frac{\int_0^\infty\big(z/y(z)\big)e^{-az^2/2}\,dz}{\int_{-\infty}^0\big(z/y(z)\big)e^{-az^2/2}\,dz}.$$ Since $z/y(z)\lessgtr 1$ when $z\gtrless 0$, we get $f(a)<1$. Also, $\lim\limits_{a\to\infty}f(a)=1$ is easy to see.