Fermi-Dirac symmetry about the Fermi level

The Fermi-Dirac function gives the probability that a state at energy $E$ is occupied:

$$f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)}$$

The probability that a state is empty is $1 - f(E)$.

Occupation of the level kT above EF

Put $E = E_F + kT$:

$$f(E_F + kT) = \frac{1}{1 + e^{1}} \quad \cdots (1)$$

Emptying of the level kT below EF

Put $E = E_F - kT$:

$$\begin{aligned} 1 - f(E_F - kT) &= 1 - \frac{1}{1 + e^{-1}} = \frac{e^{-1}}{1 + e^{-1}} \end{aligned}$$

Multiply numerator and denominator by $e$:

$$1 - f(E_F - kT) = \frac{1}{e + 1} \quad \cdots (2)$$

Comparison

(1) and (2) are identical:

$$1 - f(E_F - kT) = f(E_F + kT) = \frac{1}{1 + e} = 0.269$$

Hence the probability of a level $kT$ below $E_F$ being empty equals the probability of a level $kT$ above $E_F$ being occupied. Proved.

General result

The same steps work for any $\Delta E$:

$$1 - f(E_F - \Delta E) = \frac{1}{1 + e^{\Delta E/kT}} = f(E_F + \Delta E)$$

So $f(E)$ is symmetric about $E_F$ (about the point $f = 1/2$). Physically, the electrons that leave states just below $E_F$ are the same ones that fill states just above $E_F$; the curve drops below 1 under $E_F$ by exactly the amount it rises above 0 over $E_F$.

 f(E)
 1 |-----.
   |      \  <- holes (empty) below EF
 .5|- - - -*
   |        \ <- electrons above EF
 0 |_________'------> E
       EF-kT EF EF+kT