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 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
$$\frac{a+b}{c+1} +\frac{b+c}{a+1} +\frac{c+a}{b+1} > \frac{a+b}{a + b + c} +\frac{b+c}{a + b + c} +\frac{c+a}{a + b + c} $$
$$=  \frac{2a + 2b + 2c}{a + b + c}$$
$$ = 2$$
