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

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- Jan 28th 2013, 11:57 PMSwarnavalgebra problem
a,b>0, a+b=1, then prove that (a^2b)+(b^2a)<=1

- Jan 29th 2013, 12:23 AMtopsquarkRe: algebra problem
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 - Jan 29th 2013, 12:47 AMSwarnavRe: algebra problem
it is not (a^2)*b or(b^2)a,it is a^(2b)+b^(2a)

- Feb 7th 2013, 01:35 AMSwarnavRe: algebra problem
please someone help me to solve it.

- Feb 7th 2013, 03:53 AMpssingh1001Re: algebra problem
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 - Feb 7th 2013, 03:56 AMpssingh1001Re: algebra problem
I think this is the right ans...............

if u know something diff..............

plz share with whole team.........

Regards

prem - Feb 8th 2013, 03:04 AMCosmoBlazerRe: algebra problem
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 - Feb 15th 2013, 09:33 PMpssingh1001Re: algebra problem
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 - Feb 15th 2013, 09:35 PMpssingh1001Re: algebra problem
if u have something other easiest evaluation ..

Please share with me..........

thanks - Feb 16th 2013, 03:29 AMibduttRe: algebra problem