Least of All

Thank you Ruth from Manchester High School for Girls for your solution to this problem.

Most people's first reaction to this question is exactly the same as Ruth's:

"My initial conjecture was that the minimum value of $f(x)$ is when $x=0$ for any value of $a$, because the function is even and increases as $|x|$ increases."

Using calculus, we shall see that this is not so and that the minimum value of $f(x)$ does depend on the value of $a$ .

The given expression is $$\begin{eqnarray} f(x) &=& (1 + (a+x)^2)(1 + (a-x)^2)\\ &=&(1 + x^2 + a^2 +2ax)(1 + x^2 +a^2 -2ax)\\ &=& (1 + x^2 + a^2)^2 - 4a^2x^2 \\ &=& x^4 + 2x^2(1 + a^2) + (1 +a^2)^2 - 4a^2x^2 \\ &=& x^4 + 2x^2(1 - a^2) + (1+a^2)^2. \end{eqnarray}$$
As this is a quartic in $x$ there will be one or three turning points.

Differentiating $f$ to find the minima:
$$f'(x) = 4x^3 + 4x(1-a^2) = 4x(x^2 + (1 - a^2))$$

Case 1 : $(1 - a^2) < 0$.
The derivative $f'(x) = 0$ for $x = 0$ and $x = \pm \sqrt (a^2 - 1)$. The second derivative $f''(x) = 12x^2 + 4(1 - a^2) < 0$ at $x = 0$ which gives a maximum value but $f''(x) = 12x^2 + 4(1 - a^2) > 0$ for $x = \pm \sqrt (a^2 - 1)$ giving two minimum points on the quartic where $x = \pm \sqrt (a^2 - 1)$. The minimum value at each point is $f(x) = 4a^2$ where the position of X for these minimum values is clearly dependent of $a$.

Case 2 : $(1 - a^2) > 0$.
The derivative $f'(x) = 0$ if and only if $x = 0$. The second derivative
$f''(x) = 12x^2 + 4(1 - a^2) > 0$ so there is one minimum value $f(x)= (1+a^2)^2$ where the position of X, at $x=0$, is independent of $a$ agreeing with the conjecture.

Case 3 : $(1 - a^2) = 0$.
Note that where $a^2 = 1$ there is a single minimum $f(x) = 4$ at $x=0$ giving continuity between Case 1 and Case 2.

The most likely first conjecture agrees with Case 2 but does not allow the possibility of Case 1. Realising that the function we are minimizing is a quartic we should have taken into account the possibility of two minimum values.