how to find minimum acceptable current through voltage divider driving transistor
No, the datasheet won't give you a minimum base current, because that's determined by the collector current you want. \$H_{FE}\$ is the parameter you need. For the BC557 (no suffix) that's minimum 125, which means that for a LED current of, say, 20mA you'll have  


  \$ I_B = \dfrac{I_C}{H_{FE}} = \dfrac{20mA}{125} = 160\mu A\$  


If \$H_{FE}\$ would be higher (can be up to 800) the resistor in series with the LED will limit the current.  

You can use a divider to set the base current, but this will cause unnecessary current parallel to the base-emitter junction. Just use a series resistor:


  \$ R = \dfrac{\Delta V}{I} = \dfrac{5V - 0.7V}{160 \mu A} = 27k\Omega \$


That's the maximum resistance to get a guaranteed 20mA collector current. (I'm presuming your BC557 is connected to 5V, if it's 3.3V just redo the calculation.)  



Note that the BC557 is a PNP transistor. If you want an NPN choose the BC547.