Why do we use a current source to relax tradeoff between voltage gain and voltage headroom?
The common source amplifier has a voltage gain of $$A_v = -g_m(R_{\text{L}} \parallel r_o)$$

where \$R_{\text{L}}\$ is the load resistor, and \$r_o\$ is the small signal output resistance of the MOSFET which is generally high (usually \$R_{\text{L}} \ll r_o\$ for reasonable values of \$R_{\text{L}}\$). Increasing \$R_{\text{L}}\$ increases the gain of the circuit.

However, increasing \$R_{\text{L}}\$ also reduces the voltage headroom for the output voltage because of Ohm's Law. The output voltage in terms of \$R_{\text{L}}\$ is

$$v_{\text{out}} = V_{\text{DD}} - i_{\text{D}}R_{\text{L}}$$

However, \$v_{\text{out}}\$ is limited to the range 0V to \$V_{\text{DD}}\$ (the "headroom"). Increasing \$R_{\text{L}}\$ means that \$v_{\text{out}}\$ will hit the 0V rail sooner. You could reduce \$I_{\text{D}}\$ to improve headroom, but that also lowers the gain because \$g_m\$ is proportional to \$I_{\text{D}}\$: $$g_m = \sqrt{2\mu C_{ox}\frac{W}{L}I_{\text{D}}}$$

The only way to increase both headroom and gain with a resistively-loaded common source amplifier is to increase \$V_{\text{DD}}\$, which is usually undesirable and/or not an option.

A current source fixes this problem. An ideal current source has infinite output resistance, which would result in infinite gain for the common source. An ideal current source also supplies a constant current regardless of the voltage across it (it does not obey Ohm's Law). A real current source has finite but high output resistance and isn't entirely independent of the voltage across it, but it nonetheless relaxes this tradeoff between gain and headroom.