If the inequality 1+ _5 (x^2+1) ≥ _5 (a x^2+4 x+a) holds x ∈ R ,...

$$\begin{array}{rlrl} & \log _5\left(5 x^2+5\right) & \geq \log _5\left(a x^2+4 x+a\right) \forall x \in R \\ & \therefore & 5 x^2+5 & \geq a x^2+4 x+a \forall x \\ & \therefore & (5-a) x^2-4 x+5-a & \geq 0 \forall x \in R \\ 5-a & >0\end{array}$$ and \begin{aligned} & D \leq 0 \\ ...(2) & a < 5 \end{aligned} and \begin{aligned} 16-4(5-a)^2 & \leq 0 \\ 4-(5-a)^2 & \leq 0 \\ (a-5)^2-4 & \geq 0 \\ (a-3)(a-7) & \geq 0 \end{aligned}
From Eqns. (1) and (2) $\Rightarrow a \in(2,3]$ ...(3) $$\therefore \quad p+q=5$$