How to find the unknown in this log inequality?? Find all values of the parameter a $\in\Bbb R$ for which the following inequality is valid for all x $\in\Bbb R$.
$$ 1+\log_5(x^2+1)\ge \log_5(ax^2+4x+a) $$

I'm lost when I got to this stage: $ 5x^2-4x+5\ge ax^2+a$
I did this but still don't know how to proceed: $ (5-a)x^2-4x+(5-a)\ge0 $
My reasoning is that the discriminant for $ (5-a)x^2-4x+(5-a)\ge0 $ must be $\ge0$ because x $\in\Bbb R$. And from that I get a <= 7 or  a <= 3.
Then because $log_5(ax^2+4x+a)$, then $(ax^2+4x+a)$ > 0. 
Then..? 
Here I use same reasoning as for finding the above a <= 3 or 7 too, that is x $\in\Bbb R$. Then I get a <= -2 and a <= 2 by using discriminant. 
Where does my reasoning go wrong? Can anyone explain to me how to solve this? 
Answer given is $(2,3]$.

$\log(x)$ is monotonically increasing. Thus, assuming $ax^2+4x+a\gt0$ we can deduce
$$
\begin{array}{c}
1+\log_5(x^2+1)\ge\log_5(ax^2+4x+a)\\
\Updownarrow\\
5x^2+5\ge ax^2+4x+a\\
\Updownarrow\\
(5-a)x^2-4x+(5-a)\ge0
\end{array}\tag{1}
$$
The last inequality in $(1)$ is satisfied for $x$ on a (possibly empty) finite closed interval when $a\gt5$ and on the complement of a (possibly empty) finite open interval when $a\lt5$. We want the last inequality in $(1)$ to be satisfied for all $x\in\mathbb{R}$, so we want the solution to be the complement of an empty interval. That means we want $a\lt5$.
The endpoints of the (possibly empty) finite interval are given by
$$
x=\frac{2\pm\sqrt{-21+10a-a^2}}{5-a}\tag{2}
$$
When $3\lt a\lt7$, the discriminant is positive and so the interval is not empty. Combined with the constraint that $a\lt5$, we want $a\le3$.
We also want that $ax^2+4x+a\gt0$ for all $x$. Obviously, we need $a\gt0$. Next, we need
$$
ax^2+4x+a=a\left(x+\frac2a\right)^2+a-\frac4a\gt0\tag{3}
$$
So that $(3)$ is positive for all $x$, we need $a\gt2$.
Thus, we want $2\lt a\le3$.
