a,b>0, a+b=1, then prove that (a^2b)+(b^2a)<=1
a+ b =1 are given
then
a= 1-b
multiply a both side
a^2 = a -a*b
multiply b both side
a^2 *b = ab -a* b^2
then
a^2 *b + b^2 *a = ab
a,b>0 & a+b =1 are given
then ab>=1
then we can say that
(a^2b)+(b^2a)<=1
hence proved