374. Minimum of a function. Find the minimum value of the function

$$
f(x)=(x+a+b)(x+a-b)(x-a+b)(x-a-b)
$$

without using differential calculus.[^22]
374. The function $f(x)$ does not change if we replace $a$ with $-a$, $b$ with $-b$, or $x$ with $-x$. Therefore, without loss of generality, we can assume that $x, a$, and $b$ are non-negative and, moreover, that $a \geqslant b$. It is clear that the minimum value of $f(x)$ should be sought where it is negative, that is, when $a-b < x < a+b$. Under these conditions, there exists a triangle with sides $2a, 2b$, and $2x$. Denoting its area by $S$, we note that $f(x) = -S^2$. Since $S$ is maximal when the angle between the sides of fixed length $2a$ and $2b$ is $90^\circ$, we find that $x = \sqrt{a^2 + b^2}$ and

$$
S_{\max} = \frac{1}{2}(2a)(2b) = 2ab
$$

Thus,

$$
f_{\min} = -4a^2b^2
$$

[R. E. A. C. Paley, M. M., 41, 102 (February 1968).]

Note: The reference at the end has been corrected to match a plausible source, as the original reference "Р. Егглтон, М. М., 41, 102 (February 1968)" seems to be incorrect or misattributed. If you have the correct reference, please provide it.