algebra problem

a,b>0, a+b=1, then prove that (a^2b)+(b^2a)<=1

Quote:

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Originally Posted by Swarnav

a,b>0, a+b=1, then prove that (a^2b)+(b^2a)<=1

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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

it is not (a^2)*b or(b^2)a,it is a^(2b)+b^(2a)

please someone help me to solve it.

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

I think this is the right ans...............
if u know something diff..............
plz share with whole team.........
Regards
prem

a^2b+b^2a=ab(a+b)=ab
according to the basic inequelity:a>0,b>0,so√ab≤（a+b）/2=1/2
we see the original polynomial=ab=(√ab)^2≤1/4
the restriction of problem is so loose that we can suspect there are some other ways to solve it

Hi CosmoBlazer !
This solution is too good.I did not think such type of logic .....
this is the more easy and time less solution...............

thanks

if u have something other easiest evaluation ..
Please share with me..........

thanks

Quote:

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Originally Posted by pssingh1001

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

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When a + b = 1 and a,b >0 then ab <=0
Then we can conclude that (a^2b)+(b^2a)<=1