Minimum of which?

Suppose the minimum was greater than 0.25.

Then their sum is greater than 1.

But $a+b+c+d-a^{2}-b^{2}-c^{2}-d^{2} =-((a-0.5)^{2}+(b-0.5)^{2}+(c-0.5)^{2}+(d-0.5)^{2})+1 \leq 1$ , a contradiction. This is because all squares of reals are at least 0 by the Trivial Inequality .

The equality case for 0.25 is when $a=b=c=d=0.5$

Without loss of generality, let $a \geq\ b \geq\ c \geq\ d$ .

Therefore the minimum is $d-a^2$ .

$a \geq d$ , so let $a=d+x$ , where x is a non-negative real.

The question now reduces to finding the maximum of $f(d)=$ $d-(d+x)^2$ , which is a quadratic in d.

Note that $f(d)$ for $d=e$ , where e can be any number is maximized when $x=0$ , so let $x=0$ .

Therefore $f(d)=$ $d-d^2$ .

The maximum can then be found via a variety of ways such as the turning point formula, all of which would give the answer as $0.25$ .

Let's call $m = \textrm{min} \{a - b^2 , b - c^2, c - d^2, d - a^2 \}$ . We then of course have $a - b^2 \geq m$ $b - c^2 \geq m$ $c - d^2 \geq m$ $d - a^2 \geq m$ (and some of the inequalities MUST be an equality, but this is not crucial for the following steps).

Adding these four inequalities we have $(a+b+c + d) - (a^2 + b^2 + c^2 +d^2) \geq 4m$ that is $(-a^2 + a) + (-b^2 +b) + (-c^2 + c) + (-d^2 + d) \geq 4m$ Now it's basic that $-x^2 + x \leq \dfrac{1}{4}$ (you just have to graph the parabola and find its vertex). To get this basic result you can also complete the square $- x^2 + x = - x^2 + x - \dfrac{1}{4} + \dfrac{1}{4} = -\left(x - \dfrac{1}{2}\right)^2 + \dfrac{1}{4} \leq \dfrac{1}{4}$ where equality occurs at $x = \dfrac{1}{2}$ . So we have

$4m \leq (-a^2 + a) + (-b^2 +b) + (-c^2 + c) + (-d^2 + d) \leq \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = 1$

and we found an estimate

$m \leq \dfrac{1}{4}$

This value is in fact reached when $a=b=c=d = \dfrac{1}{2}$ , so the maximum of the expression is in indeed 0.25.

When a=b=c=d=0, the min is 0. So the max must be 0 or greater. If any of the 4 values (a,b,c,d) are not equal, then clearly the min will be negative. So it must be the case that a=b=c=d. The only way to exceed a min of 0 is when a,b,c, and d are all equal fractions. So the problem becomes: maximize x-x^2. This is a parabola of the equation y = -x^2 + x + 0. The x-coordinate of the vertex is at -b/2a = -(1)/2(-1) = -1/-2 = 1/2. Since a is -1, the parabola opens downward. The max y-coordinate is 1/2 - (1/2)^2 = 1/4 = .25.

Use min{}<AM{} and factorize the right hand side to get min>1/4 - some squares

Because four numbers: $a-b^2, b-c^2, c-d^2, d-a^2$ , make a cyclic. So the minimum can be reach when a=b=c=d. But this minimum is maximized when a, b, c, d < 1. Therefore it is when: $a = b = c = d = \frac{1}{2}, MIN=\boxed{0.25}$