Given positive real numbers $a$, $b$, $c$, $d$, $e$ with $\sum_{\text{cyc}}\,\frac{1}{4+a}=1$, prove that $\sum_{\text{cyc}}\,\frac{a}{4+a^2}\le1$.

We need to prove that $$1-\sum_{cyc}\frac{a}{4+a^2}\geq0$$ or $$\sum_{cyc}\left(\frac{1}{5}-\frac{a}{4+a^2}+3\left(\frac{1}{5}-\frac{1}{4+a}\right)\right)\geq0$$ or $$\sum_{cyc}\frac{a^3-a^2-a+1}{(4+a^2)(4+a)}\geq0.$$ Now, by AM-GM $$a^3+\frac{1}{2}\geq\frac{3}{2}a^2$$ and $$\frac{1}{2}a^2+\frac{1}{2}\geq a,$$ which after summing gives $$a^3-a^2-a+1\geq0$$ and we are done!