If  $1+{\log }_5(x^2+1)\ge {\log }_5(a{x}^2+4x+a),\forall x\in R$  then ‘a’ can equal to

we have  $1+{\log }_5(x^2+1)\ge {\log }_5(a{x}^2+4x+a)$
 $$\begin{gathered} \Rightarrow 5(x^2+1)\ge a{x}^2+4x+a,\forall x\in R \\ \Rightarrow x^2(5-a)-4x+(5-a)\ge 0\forall x\in R \end{gathered}$$
As 5>a
and  $D\le 0$ 
 $$\begin{gathered} \Rightarrow 16-4{(5-a)}^2\le 0\Rightarrow (2-(5-a))(2+5-a)\le 0 \\ \Rightarrow (a-3)(a-7)\ge 0\Rightarrow (-\infty ,3]\cap [7,\infty ) \\ \therefore \left(1\right)\cap \left(2\right)\Rightarrow a\in (-\infty ,3] \end{gathered}$$
 But for the domain of  ${\log }_5(a{x}^2+4x+a)$
We must have  $a{x}^2+4x+a>0\forall x\in R$
 $$\begin{gathered} \Rightarrow a>0and16-4a^2<0 \\ \Rightarrow a\in (-\infty ,-2)\cap (2,\infty ) \\ \Rightarrow \left(3\right)\cap \left(4\right)\Rightarrow (2,3] \end{gathered}$$
 Hence, (A) and (B) are correct.