Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality

For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} + 
\frac{7 + 2c}{1 + b} + 
\frac{7 + 2a}{1 + c} \geqslant 
\frac{69}{4}
$$

Here equality occurs for $a=b=c=\frac{1}{3}$ which is not-so-usual, so I decide to write the inequality as
$$
\frac{21 + 2q}{3 + p} + 
\frac{21 + 2r}{3 + q} + 
\frac{21 + 2p}{3 + r} \geqslant
\frac{69}{4}
$$Where the constraint now is $p + q + r = 3$ and the equality occurs for $p = q = r = 1$. Now we are left to proving that just
$$
2\sum_{cyc}{\frac{q}{3 + p}} + 
21\sum_{cyc}{\frac{1}{3 + p}} \geqslant \frac{69}{4}
$$
Now it is sufficient to prove that
$$
\sum_{cyc}\frac{q}{3 + p}\geqslant\frac{3}{4} \quad \textrm{and} \quad \sum_{cyc}\frac{1}{3 + p} \geqslant \frac{3}{4}
$$ The second is true but I can't prove that the first is true.

we have to prove $$7\sum \frac{1}{1+a}+2\sum_{cyc}\frac{b^2}{b+ab}\ge 69/4$$
but  by $AM\ge HM$ $$7\sum \frac{1}{1+a}\ge 63/4$$
also using $ab+bc+ca\le {(a+b+c)}^2/3=1/3$
$$2\sum_{cyc} \frac{b^2}{b+ab}\ge 2\frac{(a+b+c)^2}{ab+bc+ca+a+b+c}\ge 6/4$$
The conclusion is now obvious
