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?
 A: 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
$$g(t)\geq \left(a^2+1\right)\,\left(\frac{a^{a^2-1}}{a^{2a^2}}\right)^{\frac{1}{a^2+1}}=\left(a^2+1\right)\,\frac{1}{a}=g(a)\,.$$
A: For $x=0$ we get a value $\frac{7}{3}.$ 
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\frac{x^4+5x^2+7}{x^2+3}\geq\frac{7}{3}$$ or
$$x^2(3x^2+8)\geq0,$$ which is obvious. 
A: 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 $$x=0$$
A: As an alternative, using your idea for decomposition, by Rearrangement inequality with


*

*$(a_1,a_2)=(\frac13,\frac{1}{(x^2+3)})$

*$(b_1,b_2)=\left(3(x^2+3),1\right)$


we have that
$$a_1b_1+a_2b_2=\frac13\cdot 3(x^2+3)+\frac{1}{(x^2+3)}\cdot 1= x^2+3+\frac{1}{(x^2+3)}\ge a_1b_2+a_2b_1=$$
$$=\frac13\cdot 1 +\frac{1}{(x^2+3)}\cdot 3(x^2+3)=\frac13+3=\frac{10}3$$
with equality for 
$$a_1=a_2 \iff \frac13=\frac{1}{(x^2+3)}\iff x=0$$
therefore
$$f(x)=(x^2+3)+\frac{1}{x^2+3}-1\ge \frac{10}3-1=\frac 73$$
with the minimum attained at $x=0$.
A: 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
$$x^2=\frac{\frac{7}{3}-5+5-\frac{7}{3}}{2}=0$$


