Find the number of solutions to $$a-b^2\ge\frac 14,b-c^2\ge\frac 14,c-d^2\ge\frac 14,d-a^2\ge\frac 14$$
My attempt:
Adding all 4 equations, $$a+b+c+d-(a^2+b^2+c^2+d^2)\ge 1$$ $$a(1-a)+b(1-b)+c(1-c)+d(1-d)\ge1$$ Using symmetry, $$a(1-a)\ge\dfrac 14$$$$b(1-b)\ge\dfrac 14$$$$c(1-c)\ge\dfrac 14$$$$d(1-d)\ge\dfrac 14$$
Considering $a(1-a)\ge\dfrac 14$,
$$4a-4a^2\ge 1\Rightarrow 4a^2-4a+1\le 0\Rightarrow (2a-1)^2\le0$$
Thus the only value of $A$ satisfying the inequality is $a=\frac 1 2$ Similarly the only values for $b$, $c$ and $d$ are all $\frac 12$
Thus the number of solutions to the system of inequalities is 1.
$$$$While this matches the answer given, I'm not sure if my reasoning is sound.
We can add all these as $$4a-4b^2+4b-4c^2+4c-4d^2+4d-4a^2\geq 1+1+1+1$$ $$\implies (4a^2-4a+1)+(4b^2-4b+1)+(4c^2-4c+1)+(4d^2-4d+1)\leq 0$$
$$\implies (2a-1)^2+(2b-1)^2+(2c-1)^2+(2d-1)^2\leq 0.$$
We have, $$(2a-1)^2+(2b-1)^2+(2c-1)^2+(2d-1)^2=0$$
Thus, $$a=b=c=d=\frac{1}{2}$$