For $x \in \mathbb{R}$, the minimum value of $\frac{x^2+2 x+5}{x^2+4 x+10}$ is

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Aicharya · Accepted Answer

Let $\frac{x^2+2 x+5}{x^2+4 x+10}=y, y \in \mathrm{R}$ $\begin{aligned} & \Rightarrow x^2+2 x+5=y\left(x^2+4 x+10ight) \ & \Rightarrow(y-1) x^2+(4 y-2) x+(10 y-5)=0\end{aligned}$ Since, $x \in \mathbf{R}$ $\begin{aligned} & \because(4 y-2)^2-4(y-1)(10 y-5)>0 \ & \Rightarrow 6 y^2-11 y+4 \leq 0 \ & \Rightarrow(2 y-1)(3 y-4) \leq 0 \ & 	herefore y \in\left(\frac{1}{2}, \frac{4}{3}ight)\end{aligned}$ $	herefore$ Minimum value of $\frac{x^2+2 x+5}{x^2+4 x+10}$ is $\frac{1}{2}$

For $x \in \mathbb{R}$, the minimum value of $\frac{x^2+2 x+5}{x^2+4 x+10}$ is

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