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Math Help - inequality 2

  1. #1
    Member great_math's Avatar
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    inequality 2

    If a,b and c are non-negative reals such that <br />
\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge 2, prove that


    abc \le \frac{1}{8}


    thank you.
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  2. #2
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    Quote Originally Posted by great_math View Post
    If a,b and c are non-negative reals such that <br />
\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge 2. prove that abc \le \frac{1}{8}.
    let a=\tan^2x, \ b = \tan^2 y, \ c=\tan^2 z, \ \ 0 \leq x,y,z < \frac{\pi}{2}. so your inequality becomes: \cos^2x + \cos^2y+\cos^2z \ge 2. hence:

    \boxed{1} \ \ \cos^2x \ge 2 - \cos^2y - \cos^2z = \sin^2y + \sin^2z \ge 2\sin y \sin z. similarly:

    \boxed{2} \ \ \cos^2 y \ge 2\sin x \sin z,

    \boxed{3} \ \ \cos^2 z \ge 2 \sin x \sin y.

    multiplying \boxed{1} \ , \boxed{2} \ , \boxed{3} \ , together will give us: \cos^2x \cos^2y \cos^2z \ge 8 \sin^2 x \sin^2y \sin^2 z. thus: \tan^2x \tan^2y \tan^2z \le \frac{1}{8}. \ \ \ \Box
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  3. #3
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    for those who are interested, the above inequality can be easily generalized, i.e. if a_1, \cdots , a_n, \ n \geq 2, are non-negative numbers and \frac{1}{1+a_1} + \frac{1}{1+a_2}+\cdots + \frac{1}{1+a_n} \geq n-1, then:

    a_1a_2 \cdots a_n \leq \frac{1}{(n-1)^n}. the proof is the same and is based on AM-GM: let a_j=\tan^2x_j, \ 0 \leq x_j < \frac{\pi}{2}, \ 1 \leq j \leq n. then the inequality becomes: \cos^2x_1 + \cos^2x_2 + \cdots + \cos^2x_n \geq n-1. so:

    \cos^2 x_j \geq n-1 - \sum_{i \neq j} \cos^2x_i=\sum_{i \neq j} \sin^2 x_i \geq (n-1) \left(\prod_{i \neq j} \sin^2x_i \right)^{\frac{1}{n-1}}, \ \ 1 \leq j \leq n. multiplying these inequalities together gives us: \cos^2 x_1 \cos^2 x_2 \cdots \cos^2x_n \geq (n-1)^n \sin^2x_1 \sin^2x_2 \cdots \sin^2x_n.

    hence: a_1 a_2 \cdots a_n = \tan^2x_1 \tan^2x_2 \cdots \tan^2x_n \leq \frac{1}{(n-1)^n}. \ \ \ \Box
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