Paolis,
Let me first derive the formula for the 3 D case. For a free electron model, electron energy is given as ## E = \frac {\hbar^2 k^2}{2m} ##. We can invert the formula and write ## k = \sqrt {\frac {2mE}{\hbar^2}}## or ## k \sim \sqrt E##.
In 3D, the constant energy surface is a surface of a sphere. The number of electrons of energy not grater than at the surface is the reciprocal space volume of a sphere, that is ##N \sim \frac 4 3 \pi k^3 \sim E^ {\frac 3 2}##
that gives the density of states as ##g(E) = \frac {dN}{dE} \sim E^{\frac 1 2} ##
Now, the average kinetic energy is $$\left< E \right> = \frac {\int^{E_F}_0 E\cdot g(E) \, dE} {\int_0^{E_F} g(E)\, dE} = \frac {\int^{E_F}_0 E\cdot E^{\frac 1 2} \, dE} {\int_0^{E_F} E^{\frac 1 2}\, dE} = \frac 3 5 \cdot E_F$$
That's how you got the formula for the average kinetic energy in the 3-D case.
In one dimension, the number of states from zero to ##k_F## is proportional to ##k_F##, that is proportional to ##\sqrt{(E)}##. Differentiating wrt to E, we get the density of states as ## g(E) \sim E^{\frac 1 2} ##
The average kinetic energy can be calculated as before but using a 1-D density of states function, that is
$$\left< E \right> = \frac {\int^{E_F}_0 E\cdot g(E) \, dE} {\int_0^{E_F} g(E)\, dE} = \frac {\int^{E_F}_0 E\cdot E^{-\frac 1 2} \, dE} {\int_0^{E_F} E^{-\frac 1 2}\, dE} = \frac 1 3\cdot E_F$$
So, the average kinetic energy in 1-D measure relative to the bottom of the band is ## \frac 1 3 E_F ##