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Math Help - Interesting inequality

  1. #1
    MHF Contributor alexmahone's Avatar
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    Interesting inequality

    If a, b, c are real numbers such that \frac {1}{2} \le a \le 1, \frac {1}{2} \le b \le 1, \frac {1}{2} \le c \le 1;
    prove that 2 \le \frac {a + b}{1 + c} + \frac {b + c}{1 + a} + \frac {c + a}{1 + b} \le 3
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    Quote Originally Posted by alexmahone View Post
    If a, b, c are real numbers such that \frac {1}{2} \le a \le 1, \frac {1}{2} \le b \le 1, \frac {1}{2} \le c \le 1;
    prove that 2 \le \frac {a + b}{1 + c} + \frac {b + c}{1 + a} + \frac {c + a}{1 + b} \le 3
    Lemma: for any \frac{1}{2} \leq x ,y \leq 1: \ \ \frac{3y-x}{1+x}+ \frac{3x-y}{1+y} \leq 2.

    Proof: after simplifying the inequality becomes: 3x^2 - 4xy +3y^2 -2 \leq 0. for a fixed y, the quadratic function g(x)=3x^2 - 4xy + 3y^2 - 2 is convex. hence it attains its maximum at end

    points, i.e. \frac{1}{2} or 1. now 4g(1/2)=12y^2-8y-5=12y(y-1)+4y-5 < 0, because 0 < y \leq 1 < \frac{5}{4}. also: g(1)=3y^2 - 4y + 1=(y-1)(3y-1) \leq 0, because \frac{1}{3} < y \leq 1. \ \ \ \Box


    now we may assume that a \leq b \leq c. let f(a,b,c)=\frac{a+b}{1+c}+\frac{b+c}{1+a}+\frac{c+a  }{1+b}. then: f(a,b,c)=(a \ + \ b \ + \ c) \left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \right) - \left(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ri  ght). \ \ \ \ (*)

    since \frac{1}{1+c} \leq \frac{1}{1+b} \leq \frac{1}{1+a}, by Chebyshev's inequality we have: f(a,b,c) \leq 3 \left(\frac{a}{1+c}+\frac{b}{1+b}+\frac{c}{1+a} \right)-\left(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ri  ght). thus: f(a,b,c) \leq \frac{3a-c}{1+c}+\frac{3c-a}{1+a}+\frac{2b}{1+b}.

    since \frac{2b}{1+b} \leq 1, we get: f(a,b,c) \leq \frac{3a-c}{1+c} + \frac{3c-a}{1+a} + 1 \leq 3, by the Lemma. this proves the upper bound.


    proving the lower bound is much easier: since \frac{1}{1+a} \geq \frac{1}{1+b} \geq \frac{1}{1+c}, applying the second part of Chebyshev's inequality to (*), and this fact that \frac{x}{1+x} \geq \frac{1}{3}, whenever x \geq \frac{1}{2}, gives us:

    f(a,b,c) \geq 3 \left(\frac{a}{1+a} + \frac{b}{1+b}+\frac{c}{1+c} \right) - \left(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ri  ght)

    =2\left(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\  right) \geq 2 \left(\frac{1}{3} + \frac{1}{3} + \frac{1}{3} \right) = 2. \ \ \Box
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