Inequality with a+b+c=1 Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that 
$$\dfrac{bc}{a+5}+\dfrac{ca}{b+5}+\dfrac{ab}{c+5}\le \dfrac{1}{4}.$$
Progress: This is equivalent to show that 
$$\dfrac{1}{a^2+5a}+\dfrac{1}{b^2+5b}+\dfrac{1}{c^2+5c}\le \dfrac{1}{4abc}.$$
I'm not sure how to proceed further. Also equality doesn't occur when $a=b=c$, which makes me more confused. 
Edit: The original inequality is as follows:
Let $a_1,a_2,\cdots{},a_n$ be $n>2$ positive reals such that $a_1+a_2+\cdots{}+a_n=1$. Prove that,
$$\sum_{k=1}^n \dfrac{\prod\limits_{j\ne k}a_j}{a_k+n+2}\le \dfrac{1}{(n-1)^2}.$$

We'll prove a stronger inequality: $\sum\limits_{cyc}\frac{bc}{5+a}\leq\frac{1}{16}$.
Indeed, we need to prove that $$\sum\limits_{cyc}\frac{bc}{6a+5b+5c}\leq\frac{a+b+c}{16}$$ or
$$\sum\limits_{cyc}\left(\frac{a}{16}-\frac{bc}{6a+5b+5c}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{6a^2+5ab+5ac-16bc}{6a+5b+5c}\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)(3a+8c)-(c-a)(3a+8b)}{6a+5b+5c}\geq0$$ or
$$\sum\limits_{cyc}(a-b)\left(\frac{3a+8c}{6a+5b+5c}-\frac{3b+8c}{6b+5a+5c}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2(15a+15b+7c)}{(6a+5b+5c)(6b+5a+5c)}\geq0.$$
Done!
