Here's a partial proof without calculus:
Suppose ##x## has been chosen to be to be some positive value ##x_0##. From the condition
$$ 0 < y \leq 1 - x $$
the value of y is still indeterminate. However, since ##\frac{1}{y^2}## is always positive, the minimum of
$$ (1 - \frac{1}{x_0^2})(1-\frac{1}{?^2}) $$
must always occur when the maximum y "budget" is used. (The two minuses multiplied from the first and second terms will cancel in a double negative manner.)
Thus, we may reduce the possible answers to the possibilities for exactly the cases when
$$ y = 1 - x $$
without any loss of solutions.
Because the expression is symmetric about ##x## and ##1-x##, a maximum or minimum must occur at the halfway mark ##x = \frac{1}{2} ##.
$$ (1 - \frac{1}{\frac{1}{2}^2})(1-\frac{1}{\frac{1}{2}^2}) = 9 $$
Numerical checking that setting x = 0.6 and x = 0.4 gives a higher value than 9 shows that this is a minimum point, therefore:
$$ (1 - \frac{1}{x^2})(1-\frac{1}{y^2}) \geq 9 $$
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There is still the possibility however that the curve has other minimums other than on its line of symmetry:
Can it be shown without calculus that this hypothetical case is impossible?