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.
 A: 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
$$\sum \frac{7+2b}{1+a} \geq \frac{23^2}{30 + 2ab+2bc+2ca} \geq \frac{69}{4}.$$
A: 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
A: By C-S $$\sum_{cyc}\frac{7+2b}{1+a}=\sum_{cyc}\left(\frac{7+2b}{1+a}-\frac{7}{2}\right)+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{4b+7(1-a)}{1+a}+\frac{21}{2}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{11b+7c}{1+a}+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{(11b+7c)^2}{(11b+7c)(1+a)}+\frac{21}{2}\geq$$
$$\geq\frac{1}{2}\frac{324(a+b+c)^2}{\sum\limits_{cyc}(11b+7c)(1+a)}+\frac{21}{2}=\frac{162(a+b+c)^2}{18(a+b+c)^2+18(ab+ac+bc)}+\frac{21}{2}\geq$$
$$\geq\frac{162(a+b+c)^2}{18(a+b+c)^2+6(a+b+c)^2}+\frac{21}{2}=\frac{69}{4}.$$
A: Another way:flipping the inequality it suffices to prove $$\sum_{cyc}\frac{7a-2b}{1+a}\le 15/4...(3)$$.
Indeed by jensen  on $f(x)=\frac{x}{1+x}$ $$\sum_{cyc} \frac{a}{1+a}\le 3\sum_{cyc}\frac{\frac{(a+b+c)}{3}}{1+\frac{(a+b+c)}{3}}=3/4...(1)$$
This can also be done like this :
as $$\frac{x}{1+x}\le \frac{9x+1}{16}$$ which is just ${(x-1/3)}^2\ge 0$
$$\sum \frac{a}{1+a}\le \sum \frac{9a+1}{16}=3/4$$
Also similar to my previous answer (by Titu's lemma) $$\sum_{cyc}\frac{b}{1+a}=\sum \frac{b^2}{b+ab}\ge 3/4...(2)$$
By $(1)$ and $(2)$  inequality $(3)$ is completed.
