Inequality with a+b+c=1

Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that

$\frac{bc}{a+5}}+{\displaystyle \frac{ca}{b+5}}+{\displaystyle \frac{ab}{c+5}}\le {\displaystyle \frac{1}{4}}.$

Progress: This is equivalent to show that

$\frac{1}{{a}^{2}+5a}}+{\displaystyle \frac{1}{{b}^{2}+5b}}+{\displaystyle \frac{1}{{c}^{2}+5c}}\le {\displaystyle \frac{1}{4abc}}.$

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,

$\sum _{k=1}^{n}{\displaystyle \frac{\prod _{j\ne k}{a}_{j}}{{a}_{k}+n+2}}\le {\displaystyle \frac{1}{(n-1{)}^{2}}}.$

Let $a,b,c$ be positive reals such that $a+b+c=1$. Prove that

$\frac{bc}{a+5}}+{\displaystyle \frac{ca}{b+5}}+{\displaystyle \frac{ab}{c+5}}\le {\displaystyle \frac{1}{4}}.$

Progress: This is equivalent to show that

$\frac{1}{{a}^{2}+5a}}+{\displaystyle \frac{1}{{b}^{2}+5b}}+{\displaystyle \frac{1}{{c}^{2}+5c}}\le {\displaystyle \frac{1}{4abc}}.$

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,

$\sum _{k=1}^{n}{\displaystyle \frac{\prod _{j\ne k}{a}_{j}}{{a}_{k}+n+2}}\le {\displaystyle \frac{1}{(n-1{)}^{2}}}.$