Minimum value of $\frac{x^4+5x^2+7}{x^2+3}$

Question

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?

Answer 1

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 $x=0$.