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Math Help - algebra problem

  1. #1
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    algebra problem

    a,b>0, a+b=1, then prove that (a^2b)+(b^2a)<=1
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: algebra problem

    Quote Originally Posted by Swarnav View Post
    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
    a^2b + b^2 a = a^2 (1 - a) - (1 - a)^2

    Simplifying we eventually get
    a^2b + b^2 a = a^2 - a

    Now, you can probably find a more elegant way to prove that -a^2 + a \leq 1, but the simplest way to approach this is to simply graph it.

    -Dan
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  3. #3
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    Re: algebra problem

    it is not (a^2)*b or(b^2)a,it is a^(2b)+b^(2a)
    Thanks from topsquark
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    Re: algebra problem

    please someone help me to solve it.
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  5. #5
    Newbie pssingh1001's Avatar
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    Re: 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
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  6. #6
    Newbie pssingh1001's Avatar
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    Re: algebra problem

    I think this is the right ans...............
    if u know something diff..............
    plz share with whole team.........
    Regards
    prem
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  7. #7
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    Re: 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
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  8. #8
    Newbie pssingh1001's Avatar
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    Re: 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
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  9. #9
    Newbie pssingh1001's Avatar
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    Re: algebra problem

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

    thanks
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  10. #10
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    Re: algebra problem

    Quote Originally Posted by pssingh1001 View Post
    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
    When a + b = 1 and a,b >0 then ab <=0
    Then we can conclude that (a^2b)+(b^2a)<=1
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