Consider a function $F(L)=(L-L^2a)T^{L-1}$, where $0<L<\frac{1}{a}$. The constants $a$ and $T$ may take values over $]0,1[$ and $[0.01,0.1]$, respectively.
The first derivative of $F$: $\frac{dF}{dL}=(L^2+L(\frac{2}{\log(T)}-\frac{1}{a})-\frac{1}{a\log(T)}) T^{L-1}$.
$\frac{dF}{dL}=0$ $\implies$ $L_1=\frac{ (\frac{1}{a}-\frac{2}{\log(T)})- \sqrt{ (\frac{2}{\log(T)}-\frac{1}{a})^2+\frac{4}{a\log(T)} } }{ 2 }$, $L_2=\frac{ (\frac{1}{a}-\frac{2}{\log(T)})+ \sqrt{ (\frac{2}{\log(T)}-\frac{1}{a})^2+\frac{4}{a\log(T)} } }{ 2 }$.
$L_2>\frac{1}{a}$, so $L_2$ is rejected.
Is it possible to prove that $F(L)$ is concave by just saying that $F(L)$ is an increasing function for $L<L_1$ and a decreasing one for $L>L_1$?
What you state shows that $L_1$ is the maximum, but it does not imply that $F$ is concave.
In fact, $F$ is not concave on the entire interval $(0,1/a)$. Note that $$F''(1/a) = -2 \log(T) - 2a > 0,$$ since $T < e^{-a}$. So there is a point of inflection between $L_1$ and $1/a$.