How to find the minimum of $f(x)=\frac{4x^2}{\sqrt{x^2-16}}$ without using the derivative? 
Find the minimum of function $$f(x)=\frac{4x^2}{\sqrt{x^2-16}}$$ without using the derivative.

In math class we haven't learnt how to solve this kind of problems (optimization) yet. I already know that is solvable using derivatives, but there should be another way. Thanks in advance!
 A: Since the quantity inside root can't be negative for real values of x we have $$x\in (-\infty, -4]\cup [4,\infty) $$
Hence we substitute $x=4\sec \theta$ for some arbitrary $\theta$ 
Hence we need to find the minimum value of $$\frac {64\sec^2\theta}{4\tan \theta}=\frac {32}{2\sin\theta\cos\theta}$$
Now using$$2\sin\theta\cos\theta=\sin 2\theta$$
We need to find minimum value of $$\frac {32}{\sin 2\theta}$$ which is simply $32$ attained when $\sin 2\theta=1$ , i.e.$$\theta=\frac {(4n+1)\pi}{4}$$ hence at that point we have $$x=4\sec\theta=\pm 4\sqrt 2$$
A: Rewrite $f(x)=\frac {4x^2}{\sqrt{x^2-16}}=4(\sqrt{x^2-16}+\frac{16}{\sqrt{x^2-16}})$ 
Using the AM-GM inequality we get:
$$4\frac{\sqrt{x^2-16}+\frac{16}{\sqrt{x^2-16}}}{2}\geq4\times\sqrt{16}$$
then by multiplying by 2 we get:
$$f(x)\geq32$$
But is it attainable? Well we have equality when $\sqrt{x^2-16}=\frac{16}{\sqrt{x^2-16}}\implies x=\pm4\sqrt2$. And yes, it is attainable.
A: Re write as $k= (4x^2)/(x^2 -16)^.5 $,
$k^2 * ( x^2 -16) = 16x^4$
This is a quadratic in x^2. When the discriminant is 0, only one solution for x^2 is implied.As squaring creates extra solutions, check which of them is implied by your function. This is either a global minimum or maximum 
A: It is the same as finding the minimum of $\frac{4z}{\sqrt{z-16}}$ for $z>16$, or the minimum of $\frac{16 t}{\sqrt{t-1}}$ for $t>1$, or the minimum of $\frac{16(u+1)}{\sqrt{u}}$ for $u>0$, or the minimum of $16\left(v+\frac{1}{v}\right)$ for $v>0$. It is clearly $\color{red}{32}$ by the AM-GM inequality.
A: Substitution makes it easier: Write $t=x^2-16$ then using inequality between AG and GM we have $$\frac{4x^2}{\sqrt{x^2-16}}=\frac{4t+64}{\sqrt{t}}\geq {4\sqrt{16t}\over \sqrt{t}}=16$$
and the minimum is achieved at $t=16$ or $x =\pm 4\sqrt{2}$
A: The solution can be arrived at by using the quadratic formula.
To begin, we want to analyze how $f(x)=\frac{4x^2}{\sqrt{x^2-16}}$ on the domain $x \gt 4$ can take on any value $h \gt 0$:
$\tag 1 f(x) = h \text{ iff } 16 x^4 - h^2 x^2 + 16 h^2 = 0$
For this quadratic equation in $x^2$ to have any solutions the discriminant must be greater than or equal to zero:
$\tag 2 h^4 - (4)(16)(16 h^2) \ge 0 \text{ iff }  h^2 \ge (4)(16)(16)  \text{ iff } h \ge (2)(16)  \text{ iff } h \ge 32$.
So the range of the function is the interval $[32, +\infty)$ and so the minimum value hit by $f$ is $32$.
Using $\text{(1)}$ and $\text{(2)}$ above and the quadratic formula when the discriminant is zero, we see that
$\tag 3 x^2 = \text{[negative } b \text{ divided by } 2 a \text{]} = (32^2)/(2)(16) \text{ iff } x = 32/(4 \sqrt 2) = 4 \sqrt 2$ so 
$\tag 4 f(4 \sqrt 2) = 32$

See Ethan Horsfall's answer.
A: Render
$\frac{4x^2}{\sqrt{x^2-16}}=a$
Here, $a$ is the minimum function value.
Square both sides and convert the resulting polynomial relation to the form $P(x)=0$:
$\frac{16x^4}{x^2-16}=a^2$
$x^4-a^2x^2+16a^2=0$
Now if $a$ is to be the minimum function for $f(x)$ value then the polynomial will have a squared factor for that value of $a$; the value of $x$ where that occurs (called $x_0$) is the resulting double root:
$x^4-a^2x^2+16a^2=(x-x_0)^2(x^2+px+q)=0$
Then
$x^4-a^2x^2+16a^2=(x-x_0)^2(x^2+px+q)=x^4+(p-2x_0)x^3+(q-2px_0+x_0^2)x^2+(-2qx_0+px_0^2)x+qx_0^2$
For this factorization to hold The cubic and linear terms must satisfy $p-2x_0=0$ and $-2q+px_0=0$ (the function would not be defined at $x_0=0$). Therefore $p=2x_0, q=x_0^2$.  Put these into the quadratic and constant terms:
$q-2px_0+x_0^2=-2x_0^2=-a^2$
$qx_0^2=x_0^4=16a^2$
Dividing the second of these two equations by the first leads to $x_0^2=32$, thus $x_0=\pm 4\sqrt{2}$ for the minimizing $x$ value.  The minimum function value, which is clearly positive, is then $a=\sqrt(2x_0^2)=8$.
A: A bit of trickery:
Note :$|x|>4 .$
Set $y= (x^2-16)^{1/2}$, $y>0$.
With $x^2= y^2 +16$:
$F(y): =\dfrac {4y^2 +64}{y}=$
$4y +64/y =$
$4(√y -4/√y)^2+ (4)(8)$.
$F_{min}(y)= 32$; at $y= 4.$
Corresponding $x$:
$y^2+16 = x^2,$ or 
$x= ^+_-(32)^{1/2}$
