Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$ Minimum value of $$f(x)=\frac{x^4+5x^2+7}{x^2+3}$$
we have $f(x)$ as 
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1$$
Now by $AM \gt GM$ we have
$$(x^2+3)+\frac{1}{x^2+3} \gt 2$$
But equality cannot occur since $$x^2+3 \ne \frac{1}{x^2+3}$$
But my question is without using calculus is there any way to find minimum using AM, GM?

Let $y=f(x)$, then
\begin{align}
  0 &= x^4+(5-y)x^2+(7-3y) \\
  x^2 &= \frac{y-5 \pm \sqrt{(5-y)^2+4(\color{red}{3y-7})}}{2} \\
  &= \frac{y-5 \pm \sqrt{y^2+2y-3}}{2} \\
  &= \frac{y-5 \pm \sqrt{(y+3)(y-1)}}{2}
\end{align}
Now,
$$\Delta=(y+3)(y-1) \ge 0$$
Together with $f(x)>0$,
$$y \ge 1$$
in which $\Delta$ is increasing with $y$.


*

*When $y \ge 5$,


\begin{align}
  3y-7 & \ge 3(5)-7 \\
  & = 8 \\
  & > 0 \\
  \Delta & > \sqrt{(5-y)^2} \\
  & = y-5 \\
\end{align}


*

*When $1 \le y \le 5$,


\begin{align}
  y-5+\sqrt{(y+3)(y-1)} & \ge 0 \tag{$x^2 \ge 0$} \\
  (y+3)(y-1) & \ge (5-y)^2 \\
  4(\color{red}{3y-7}) & \ge 0 \\
  y & \ge \frac{7}{3} \\ 
\end{align}
The non-negativity of $x^2$ imposes $$x^2=\frac{y-5+\sqrt{(y+3)(y-1)}}{2}$$
which is increasing with $y$.

The required minimum is $\frac{7}{3}$ which is achieved when
$$x^2=\frac{\frac{7}{3}-5+5-\frac{7}{3}}{2}=0$$


