ya..nice question
u can do it for a current I splitting into two parts and going to say R1 (x current) and R2( I-x current)
R1 and R2 are resistances..hmm now do one thing
P(x) = x^2(R1) + (I-x)^2 (R2)
differentiate to get the minimum, => x= R2(I)/ (R1+R2)
and what u see??? u get the expected value of x( current gets divided in the two branches in inverse ratio of R1 and R2 :) )
this completes the proof