Prove inequality $2\le\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} \le3$ 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 :) 
