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!

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$\frac{4x^2}{\sqrt{x^2-16}}=a$
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 $a=\sqrt(2x_0^2)=8$.
