<s>[INST] 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.
 [/INST] 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.
</s>