Here is a possible approach, more ad hoc than those previously suggested. Let $E=E(f,L,a)$ be the expression without the "manifestly positive" factor that Willie Wong noticed is irrelevant. Hope that establishing it for integers $a$ will lead to settling it for real $a$ (that's just a hope). So focus on integral $a$. Because $a=1$ is a bit different, separate that case off. So now explore $E(f,L,a)$ for $1 < a \le L-2$, where both $a$ and $L$ are integers. For $L$ even, $$E = -2^a \; f^{a+1} \; (1+f)^{a+1} \; \mathrm{poly}(f^{L+1}),$$ where $\mathrm{poly}(f^{L+1})$ is a polynomial in $f$ of degree $L+1$. For $L$ odd, $$E = -2^a \; f^a \; (1+f)^{a+2} \; \mathrm{poly}(f^L).$$ Examples, $L$ even: $$L=6,a=2: \quad E = -8 f^2 (f+1)^3 \left(34 f^7-31 f^6-56 f^5+59 f^4+10 f^3+23 f^2-20 f-19\right).$$ $$L=6,a=3: \quad E = -16 f^3 (f+1)^4 \left(46 f^7-27 f^6-76 f^5+19 f^4+110 f^3-5 f^2-48 f-19\right).$$ $$L=6,a=4: \quad E = -32 f^4 (f+1)^5 \left(44 f^7-15 f^6-90 f^5+55 f^4+80 f^3+15 f^2-66 f-23\right).$$
$L$ odd: $$L=7,a=2: \quad E = -8 f^2 (f+1)^4 \left(74 f^7-127 f^6+91 f^4-46 f^3+39 f^2+4 f-35\right).$$
$$L=7,a=3: \quad E = -16 f^3 (f+1)^5 \left(106 f^7-147 f^6-36 f^5+55 f^4+74 f^3+27 f^2-48 f-31\right).$$
$$L=7,a=4: \quad E = -32 f^4 (f+1)^6 \left(118 f^7-143 f^6-56 f^5+15 f^4+174 f^3-f^2-76 f-31\right).$$
$$L=7,a=5: \quad E = -64 f^5 (f+1)^7 \left(104 f^7-105 f^6-106 f^5+137 f^4+28 f^3+73 f^2-90 f-41\right).$$
Now the task is prove that $\mathrm{poly}(\;)$ is negative for $f$ in $[0,1]$. As observed previously, $f=1$ is a root, so $(f-1)$ is a factor. Just taking the last polynomial above as an example, it has a root at $f=-0.346213$ and is negative between there and $f=1$. It seems feasible to analyze the structure of $\mathrm{poly}(\;)$ and prove that it has no roots in $[0,1]$, which would settle it for integers $a>1$.
Of course I am aware that I am leaving much to hope and further work.
