<s>[INST] 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?
 [/INST] Hint: We have
$$\frac{x^4+5x^2+7}{x^2+3}\geq 7/3$$ this is equivalent to
$$3x^4+15x^2+21\geq 7x^2+21$$
$$x^2(3x^2+8)\geq 0$$ the equal sign holds it $$x=0$$
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