Interesting inequality

alexmahone · #1 $Old$ 10-17-2008, 01:07 AM

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 $2 \le \frac {a + b}{1 + c} + \frac {b + c}{1 + a} + \frac {c + a}{1 + b} \le 3$

NonCommAlg · #2 $Old$ 10-18-2008, 06:36 PM

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: $f(a,b,c)=(a \ + \ b \ + \ c) \left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \right) - \left(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\right). \ \ \ \ (*)$

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

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

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