1. ## Interesting inequality

If $\displaystyle a, b, c$ are real numbers such that $\displaystyle \frac {1}{2} \le a \le 1, \frac {1}{2} \le b \le 1, \frac {1}{2} \le c \le 1$;
prove that $\displaystyle 2 \le \frac {a + b}{1 + c} + \frac {b + c}{1 + a} + \frac {c + a}{1 + b} \le 3$

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

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

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

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

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

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

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

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

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