How do we minimize $\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right)$? Find the minimum value of the following function, where a and b are real
numbers.
\begin{align} f(x)&=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) \end{align}
Note: The solution should not contain any calculus methods.
So I conjugated the parenthesizes two by two, and I got:
\begin{align} f(x)&=\left(x^2-(a+b)^2\right)\left(x^2-(a-b)^2\right) \end{align}
Can somebody help me to take this problem further?

By your work
$$f(x)=x^4-2(a^2+b^2)x^2+(a^2-b^2)^2$$ or
$$f(x)=(x^2-a^2-b^2)^2-(a^2+b^2)^2+(a^2-b^2)^2$$ or
$$f(x)=(x^2-a^2-b^2)^2-4a^2b^2.$$
Now, we see that
$$f(x)\geq-4a^2b^2$$ and the equality occurs for $x^2=a^2+b^2$, which is possible.
Thus, $$\min_{\mathbb R}f=-4a^2b^2.$$
Done!
