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.