Assume that $a,b,c$ are real numbers from the interval $(\frac{1}{2},1)$. What is the proof that
$2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$ holds?
Assume that $a,b,c$ are real numbers from the interval $(\frac{1}{2},1)$. What is the proof that
$2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$ holds?
For the first inequality
Since the sum is symmetric, without loss of generality we can assume $a \leq b \leq c$.
Then
$\frac{a+b}{c+1} \leq \frac{c+a}{b+1} \le \frac{b+c}{a+1} $ $c+1 \geq b+1 \geq a+1$
Then, by Chebyshev's sum inequality we have
$ 3( a+b+a+c+b+c) \leq \left( \frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \right) \left( a+1+b+1+c+1 \right)$
Thus
$\left( \frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \right) \geq \frac{6(a+b+c)}{a+b+c+3}$
It is easy to see that $\frac{6(a+b+c)}{a+b+c+3} \geq 2 \Leftrightarrow a+b+c \geq \frac{3}{2}$
Second Inequality This is equivalent to
$\sum (a+b)(a+1)(b+1) \leq 3 (a+1)(b+1)(c+1)$
Again, without loss of generality we can assume that $a \leq b \leq c$.
Then
$(a+b)(a+1)(b+1) =(a+b)(ab+a+b+1)=a^2b+ab^2+a^2+b^2+2ab+a+b$
Thus, the inequality becomes
$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2+2ab+2ac+2bc+2a+2b+2c \leq $ $\leq 3abc+3ab+3ac+3bc+3a+3b+3c+3$
Or
$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2 \leq $ $\leq 3abc+ab+ac+bc+a+b+c+3 (*)$
We start by observing that $(a+1)(1-b)(1-c) \geq 0$ thus
$abc+a+bc +1 \geq ab+ac+b+c$ Similarly $abc+b+ac +1 \geq bc+ab+a+b$ $abc+c+ab +1 \geq ac+bc+a+b$
Adding them we get
$ab+ac+bc+a+b+c \leq 3abc+3$
Thus, to prove $(*)$ it is enough to show that
$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2a^2+2b^2+2c^2 \leq $ $\leq 2(ab+ac+bc+a+b+c) $
which follows immediately from $a\leq 1 \Rightarrow a^2b \leq ab$ and similar ones.
*P.S. * This Solution is terrible, there should be a much simpler one. The inequality $(a+1)(1-b)(1-c) \geq 0$ I used suggests that the Schur Inequality should be the Key, but I couldn't find it :)
For the left inequality: Since the sum of any two of $a,b,c$ is greater than 1, one has that $a+1$, $b+1$, $c + 1 < a + b + c$. So you have $\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} > \frac{a+b}{a + b + c} +\frac{b+c}{a + b + c} +\frac{c+a}{a + b + c} $ $= \frac{2a + 2b + 2c}{a + b + c}$ $ = 2$
Let $f(a,b,c)=\sum\limits_{cyc}\frac{a+b}{c+1}$.
Hence, $\frac{\partial^2f}{\partial a^2}=\frac{2(b+c)}{(a+1)^3}>0$, which says that $f$ is a convex function of $a$, of $b$ and of $c$.
Thus, $\max_{\{a,b,c\}\subset\left[\frac{1}{2},1\right]}f=\max_{\{a,b,c\}\subset\{\frac{1}{2},1\}}f=f(1,1,1)=3.$ Done!