How do we minimize $\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right)$?

Question

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?

Answer 1

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!

Answer 2

Hint:

$$(y-2c)(y-2d)=(y-(c+d))^2+4cd-(c+d)^2\ge?$$

Answer 3

Just another solution in the same spirit as lab bhattacharjee's answer.

$$ f(x)=\left(x+a+b\right)\left(x+a-b\right)\left(x-a+b\right)\left(x-a-b\right) $$ $$f(x)=x^4-2\left( a^2+1\right) x^2+(a^2-1)^2$$ Complete the square $$f(x)=\left(x^2-(a^2+1)\right)^2-(a^2+1)^2+(a^2-1)^2=\left(x^2-(a^2+1)\right)^2-4a^2$$