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.

Titu's lemma gives us

 $$ \sum \frac{q}{ 3 + p } = \sum \frac{ q^2 } { 3q + pq} \geq  \frac{ (p+q+r)^2 } { \sum 3q + pq } = \frac{ 9}{9 + pq+qr+rs}.$$

Since $(p+q+r) ^2 \geq 3 (pq+qr+rs)$, so $3 \geq pq+qr+rs $ and thus
$$ \sum \frac{q}{ 3 + p } \geq \frac{9}{9 + pq+qr+rs} \geq  \frac{3}{4}.$$

The original problem could be approach in as similar manner.

 $$\sum \frac{7+2b}{1+a} = \sum \frac{ (7+2b)^2}{(7+2b)(1+a)} \geq \frac{ (21 + 2a+2b+2c)^2}{ \sum 7 + 2b + 7a + 2ab} = \frac{23^2}{30 + 2ab+2bc+2ca}.  $$

Then, since $ \frac{1}{3} \geq ab+bc+ca$, thus
$$\sum \frac{7+2b}{1+a} \geq \frac{23^2}{30 + 2ab+2bc+2ca} \geq \frac{69}{4}.$$
