Given $7+2 b=5+2(1+b)$, we write the left-hand side in the form

$$5\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)+2\left(\frac{1+b}{1+a}+\frac{1+c}{1+b}+\frac{1+a}{1+c}\right)$$

Using the inequality $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geqslant \frac{9}{x+y+z}$, we bound the first term from below by $\frac{45}{4}$.
Using the inequality $x+y+z \geqslant 3 \sqrt[3]{x y z}$, we bound the second term from below by 6. The assertion to be proved follows.