Find $\min$ for$ f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b)$

Question

I'm trying to find $minf(x)$ for $f(x) := (x + a + b)(x + a - b)(x - a + b)(x - a - b)$, where $a, b \in \mathbb{R},$ using inequalities. For example, i can find $maxf(x)$, using AM-GM ineq:
$$\sqrt[4]{(x + a + b)(x + a - b)(x - a + b)(x - a - b)})^4 \leq \Big (\frac{x + a + b + x + a - b+ x - a + b+ x - a - b}{4}\Big)^4 = $$ $$= x^4.$$ So $maxf(x) = x^4$. But i don't know how to solve is for $minf(x)$, which i need to find.

Sure do i can find structure of square difference in $f(x)$ and we can rewrite our equality: $$f(x) = (x^2 - (a + b)^2) (x^2 - (b - a)^2).$$ But i don't know what to do next.

UPD: We need to find extremum on $x$ via fixing $a, b$. I understand that $max$ is found wrong way. How can i do it correctly?

Answer 1

$$f(x)=(x^2-(a^2+b^2))^2+(b^2-a^2)^2-(a^2+b^2)^2\ge(b^2-a^2)^2-(a^2+b^2)^2=-4a^2b^2$$ The equality occurs for $x^2=a^2+b^2$, which is always possible.

Answer 2

Here is another approach:

Denote $t=a+b$ and $s=a-b$, then $f(x)$ becomes to be $(x^2-t^2)(x^2-s^2)$. Henece $f(x)$ is quintic which is very easy to work with, more specifically, $$f(x)=x^4-(s^2+t^2)x^2+t^2s^2.$$ Now, you can make another substitution, namely $y=x^2$ to obtain a quadratic, which you know how to find a minimum of it. Don't forget to returnt to given variables.

Answer 3

For a fixed nonzero $x,$ there is no minimum and no maximum.

To exceed any upper bound, take $a = t$ and $b=0$ for some large $t.$ The original polynomial becomes $t^4 - 2 x^2 t^2 = t^2 (t^2 - 2 x^2) \; . \;$ As soon as $t^2 > 2 x^2 + 1,$ the polynomial is larger than $t^2,$ and we may take $t$ as large as we like. In particular, if $t^2 > 3 x^2,$ the value is bigger than $3 x^4$

For negative, take $a=b=t,$ gives $-4x^2 t^2 + x^4.$