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

My question is how to prove this inequality by using AM-GM inequality?
My solution (using the Chebyshev inequality).
Since $\dfrac{1}{4+a}+\dfrac{1}{4+b}+\dfrac{1}{4+c}+\dfrac{1}{4+d}+\dfrac{1}{4+e} =1,$ we have
$$1 = \dfrac{1}{4+a}+\dfrac{1}{4+b}+\dfrac{1}{4+c}+\dfrac{1}{4+d}+\dfrac{1}{4+e} \geq \dfrac{a}{4+a^2}+\dfrac{b}{4+b^2}+\dfrac{c}{4+c^2}+\dfrac{d}{4+d^2}+\dfrac{e}{4+e^2}$$
$$\Leftrightarrow \dfrac{1-a}{(4+a)(4+a^2)}+\dfrac{1-b}{(4+b)(4+b^2)}+\dfrac{1-c}{(4+c)(4+c^2)}+\dfrac{1-d}{(4+d)(4+d^2)}+\dfrac{1-e}{(4+e)(4+e^2)} \geq 0.$$
Suppose that $a \geq b \geq c \geq d \geq e$. Then, we get
$$\dfrac{1-a}{4+a} \leq \dfrac{1-b}{4+b} \leq \dfrac{1-c}{4+c} \leq \dfrac{1-d}{4+d} \leq \dfrac{1-e}{4+e}.$$
and $$\dfrac{1}{4+a^2} \leq \dfrac{1}{4+b^2} \leq \dfrac{1}{4+c^2} \leq \dfrac{1}{4+d^2} \leq \dfrac{1}{4+e^2}.$$
Applying the Chebyshev inequality, one gets
$$ \sum_{cyc}\dfrac{1-a}{(4+a)(4+a^2)} \geq \dfrac{1}{5} \sum_{cyc}\dfrac{1-a}{4+a}. \sum_{cyc}\dfrac{1}{4+a^2} = \dfrac{1}{5}\sum_{cyc}\dfrac{1}{4+a^2} \sum_{cyc} \left( \dfrac{5}{4+a}-1 \right)=0.$$

Another way.
By AM-GM $$\sum_{cyc}\frac{a}{4+a^2}\leq\sum_{cyc}\frac{a}{2a+3}.$$
Thus, it's enough to prove that:
$$1-\sum_{cyc}\frac{a}{2a+3}\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{a}{2a+3}\right)\geq0$$ or $$\sum_{cyc}\frac{1-a}{2a+3}\geq0$$ or
$$\sum_{cyc}\left(\frac{1-a}{2a+3}+5\left(\frac{1}{5}-\frac{1}{4+a}\right)\right)\geq0$$  or
$$\sum_{cyc}\frac{(a-1)^2}{(2a+3)(4+a)}\geq0.$$
