minmax it

We can assume that M is the maximum of the two expressions....
So, we know that
M >= a^2 + b
M >= b^2 + a
Adding these two gives
2M >= (a+1/2)^2 +(b+1/2)^2 -1/2
Hence we find
M >= -0.25 !!
This occurs when a=b=-0.5

The above minimax function is minimized when BOTH arguments equal each other, hence:

$a^2 + b = b^2 + a \Rightarrow a^2 - b^2 + b - a = 0 \Rightarrow (a+b)(a-b) - (a-b) = 0 \Rightarrow (a-b)(a + b - 1) = 0$ ;

or $a = b$ (i) and $a + b = 1$ (ii). Now, substitution of (i) into either argument yields the function $f(a) = a^2 + a$ , which has a critical point at:

$f'(a) = 2a + 1 = 0 \Rightarrow a = -\frac{1}{2}$ and achieves a minimum at this critical point due to $f''(-1/2) = 2 > 0$ . Now, applying (ii) to either argument yields $g(a) = a^2 - a + 1$ , where $g'(a) = 2a - 1 = 0 \Rightarrow a = \frac{1}{2}$ and $g''(1/2) = 2 > 0$ . Hence our two critical pairs come to $(a,b) = (-1/2, -1/2); (1/2, 1/2).$

Substitution of these critical pairs into the original minimax function produces:

$(a,b) = (-1/2, -1/2) \Rightarrow (-1/2)^2 - 1/2 = -\frac{1}{4}$

$(a,b) = (1/2, 1/2) \Rightarrow (1/2)^2 + 1/2 = \frac{3}{4}$

which gives a minimum value of $\boxed{-\frac{1}{4}}.$