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 $x=0$.

Answer 2

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 $$g(t)-g(a)=(t-a)\left(1-\frac{1}{at}\right)\,.$$

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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 $$g(t)\geq \left(a^2+1\right)\,\left(\frac{a^{a^2-1}}{a^{2a^2}}\right)^{\frac{1}{a^2+1}}=\left(a^2+1\right)\,\frac{1}{a}=g(a)\,.$$

Answer 3

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.

Answer 4

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 $$x=0$$

Answer 5

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

$$x^2=\frac{\frac{7}{3}-5+5-\frac{7}{3}}{2}=0$$