a,b>0, a+b=1, then prove that (a^2b)+(b^2a)<=1
If a + b = 1 then we know that b = 1 - a. So we have
$\displaystyle a^2b + b^2 a = a^2 (1 - a) - (1 - a)^2 $
Simplifying we eventually get
$\displaystyle a^2b + b^2 a = a^2 - a$
Now, you can probably find a more elegant way to prove that $\displaystyle -a^2 + a \leq 1$, but the simplest way to approach this is to simply graph it.
-Dan
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