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Math Help - prove that a^2 + b^2 + c^2 + 2abc <2

  1. #1
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    prove that a^2 + b^2 + c^2 + 2abc <2

    for a,b,c real positive numbers such that a+b+c=2, a<b+c, b< c+a , c< a+b; show that

    a^2 + b^2 + c^2 + 2abc < 2 .
    first, i try to find abc and i got
    abc< abc+a(c^2)+(a^2)b+c(b^2)+(c^2)b+ a(b^2)+abc
    then i find (a^2)+(b^2)+(c^2)< 2(a^2)+2(b^2)+2(c^2)+2(ab+ac+bc)

    i add both 2abc+(a^2)+(b^2)+(c^2) but i didnt end up with proved..
    what can i do?
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  2. #2
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    Quote Originally Posted by nh149 View Post
    for a,b,c real positive numbers such that a+b+c=2, a<b+c, b< c+a , c< a+b; show that

    a^2 + b^2 + c^2 + 2abc < 2 .
    We want to prove

    (1)\qquad a^2 + b^2 + c^2 + 2abc < 2.

    Multiply both sides by 4:

    (2)\qquad 4(a^2 + b^2 + c^2) + 8abc < 8.

    Since a+b+c=2, we can make (2) homogenous of degree three by writing it as

    (3)\qquad 2(a+b+c)(a^2 + b^2 + c^2) + 8abc < (a+b+c)^3.

    Multiply out the brackets in (3) and rearrange it a bit, and you will get

    (4)\qquad a^3+b^3+c^3 +2abc <a^2(b+c) + b^2(c+a) + c^2(a+b).

    So far, the argument is reversible, so if we can prove (4) then (1) will also hold.

    We are told that a<b+c, b< c+a and c< a+b. Therefore

    (5)\qquad (b+c-a)(c+a-b)(a+b-c) > 0.

    Multiply out the brackets in (5), rearrange it a bit, and you get exactly the inequality (4), as wanted.
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Nice proof, Opalg.

    Here is another proof (which I took all night thinking about).

    a\ <\ b+c
    \implies\ a\ <\ a+b+c-a\ =\ 2-a

    \implies\ 1-a\ >\ 0

    Similarly 1-b>0 and 1-c>0.

    \therefore\ (1-a)(1-b)(1-c)\ >\ 0

    \implies\ 1-(a+b+c)+ab+bc+ca-abc\ >\ 0

    \implies\ ab+bc+ca-abc\ >\ 1

    Now multiply through by 2, replace 2(ab+bc+ca) by (a+b+c)^2-(a^2+b^2+c^2) and rearrange to get the desired inequality.
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  4. #4
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    Quote Originally Posted by TheAbstractionist View Post
    Nice proof, Opalg.

    Here is another proof (which I took all night thinking about).
    It kept me awake for quite a while too!
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