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Math Help - Inequality

  1. #1
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    Inequality

    Given that a+b\le c+1, b+c \le a+1, c+a\le b+1 for the numbers a,b,c \ge0, show that a^2+b^2+c^2\le2abc+1.
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  2. #2
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    I am interested in this question. Squaring out some terms, I get the following inequality:

    2(ab+ac+bc)-3 \le a^2 + b^2 + c^2

    Definitely not what we are looking for!!
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  3. #3
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    Quote Originally Posted by colby2152 View Post
    I am interested in this question. Squaring out some terms, I get the following inequality:

    2(ab+ac+bc)-3 \le a^2 + b^2 + c^2

    Definitely not what we are looking for!!
    Do you want a solution?

    We deduce the problem is equivalent to
    (a+b+c+1)^2  \leq 2(1+a)(1+c)(1+b)

    This is a much better form
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  4. #4
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    Notice that the inequality we need to prove contains a term in abc. Thus we have to try somehow or other to bring that term into our working.

    Now, since we are working with non-negative numbers, we can multiply all the inequalities together. Letís try that.

    (a+b)(b+c)(c+a)\ \leq\ (a+1)(b+1)(c+1)

    The LHS is equal to (ab+bc+ca)(a+b+c)-abc (expand them out and check). The RHS is equal to abc+(ab+bc+ca)+(a+b+c)+1. Hence

    (ab+bc+ca)(a+b+c)\ \leq\ 2abc+(ab+bc+ca)+(a+b+c)+1

    Which doesnít bring us any closer to a solution, but at least we now have the abc term to work with.
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  5. #5
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    I'm sure this is supposed to be a slick application of one of those Jensen-type inequalities that they drill math olympiad candidates with. Here's a much more pedestrian approach (aesthetically unappealing because it breaks the symmetry).

    If a+b\leqslant c+1 and c+a\leqslant b+1 then b-c\leqslant 1-a and c-b\leqslant 1-a. Thus |b-c|\leqslant 1-a. In particular, 1-a\geqslant0. Also, (b-c)^2\leqslant (1-a)^2. Then

    . . . . . . \begin{array}{rcl}a^2+b^2+c^2 -2abc &=& <br />
a^2 + \frac12(1+a)(b-c)^2 + \frac12(1-a)(b+c)^2 \vspace{1ex} \\<br />
&\leqslant&a^2 + \frac12((1+a)(1-a)^2 + \frac12(1-a)(1+a)^2 \vspace{1ex} \\ &=&a^2 + \frac12((1-a^2)(1-a) + \frac12(1-a^2)(1+a) \vspace{1ex} \\<br />
&=& a^2 + 1 - a^2 = 1.\end{array}
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  6. #6
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    Quote Originally Posted by Opalg View Post
    I'm sure this is supposed to be a slick application of one of those Jensen-type inequalities that they drill math olympiad candidates with.
    Using Jensen's inequality is recent.
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  7. #7
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    Quote Originally Posted by Opalg View Post
    I'm sure this is supposed to be a slick application of one of those Jensen-type inequalities that they drill math olympiad candidates with. Here's a much more pedestrian approach (aesthetically unappealing because it breaks the symmetry).

    If a+b\leqslant c+1 and c+a\leqslant b+1 then b-c\leqslant 1-a and c-b\leqslant 1-a. Thus |b-c|\leqslant 1-a. In particular, 1-a\geqslant0. Also, (b-c)^2\leqslant (1-a)^2. Then

    . . . . . . \begin{array}{rcl}a^2+b^2+c^2 -2abc &=& <br />
a^2 + \frac12(1+a)(b-c)^2 + \frac12(1-a)(b+c)^2 \vspace{1ex} \\<br />
&\leqslant&a^2 + \frac12((1+a)(1-a)^2 + \frac12(1-a)(1+a)^2 \vspace{1ex} \\ &=&a^2 + \frac12((1-a^2)(1-a) + \frac12(1-a^2)(1+a) \vspace{1ex} \\<br />
&=& a^2 + 1 - a^2 = 1.\end{array}
    That is a brilliant solution, Opalg
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