
for q.19
since all terms of second bracket are in squares
and if i want to minimize them, one of em i took them to be equal.
let c=a
then i can write (2a+b)(ab)^2
again, minimizing them...the minimum value (ab)^2 with strictly positive value can be 1
therefore, b=2, a=c=1
put into 2a+b
u get 4
u can try in a more noble way, which is pure mathematics, but rite now neither i have that much time nor i would be having during xam
so u knoe what will be our best strategy :P
for q.20
with no condition on a, b, c
try a=1, b=c=0
nothing showed up
bingo
D
:D
