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.

By C-S $$\sum_{cyc}\frac{7+2b}{1+a}=\sum_{cyc}\left(\frac{7+2b}{1+a}-\frac{7}{2}\right)+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{4b+7(1-a)}{1+a}+\frac{21}{2}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{11b+7c}{1+a}+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{(11b+7c)^2}{(11b+7c)(1+a)}+\frac{21}{2}\geq$$
$$\geq\frac{1}{2}\frac{324(a+b+c)^2}{\sum\limits_{cyc}(11b+7c)(1+a)}+\frac{21}{2}=\frac{162(a+b+c)^2}{18(a+b+c)^2+18(ab+ac+bc)}+\frac{21}{2}\geq$$
$$\geq\frac{162(a+b+c)^2}{18(a+b+c)^2+6(a+b+c)^2}+\frac{21}{2}=\frac{69}{4}.$$
