<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] Let $a\geq 1$ be fixed.  Then, $g:[a,\infty)\to\mathbb{R}$ defined by $$g(t):=t+\frac{1}{t}\text{ for all }t\in[a,\infty)$$ is minimized at $t=a$.  To show this, write
$$g(t)-g(a)=(t-a)\left(1-\frac{1}{at}\right)\,.$$

With the Weighted AM-GM inequality, we note that
$$g(t)=a^2\,\left(\frac{t}{a^2}\right)+\frac{1}{t}\geq \left(a^2+1\right)\left(\left(\frac{t}{a^2}\right)^{a^2}\,\frac{1}{t}\right)^{\frac{1}{a^2+1}}=\left(a^2+1\right)\,\left(\frac{t^{a^2-1}}{a^{2a^2}}\right)^{\frac{1}{a^2+1}}\,\,.$$
Since $t\geq a\geq1$, we get
$$g(t)\geq \left(a^2+1\right)\,\left(\frac{a^{a^2-1}}{a^{2a^2}}\right)^{\frac{1}{a^2+1}}=\left(a^2+1\right)\,\frac{1}{a}=g(a)\,.$$
</s>