Indian Statistical Institute B.Math & B.Stat : Square of a real is non-negative

Show that the following system of inequalities has exactly one solution $a-b^2 \geq \frac{1}{4},$ $b-c^2 \geq \frac{1}{4},$ $c-d^2 \geq \frac{1}{4}$ and $d-a^2 \geq \frac{1}{4}.$ Adding up all the inequalities we get $a-b^2 + b-c^2 + c-d^2 + d-a^2 \geq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} +\frac{1}{4}$ $\implies a-a^2 - \frac{1}{4} + b-b^2- \frac{1}{4} + c-c^2 - \frac{1}{4} + d-d^2 - \frac{1}{4} \geq 0$ $\implies -\big(a- \frac{1}{2} \big)^2 -\big(b- \frac{1}{2} \big)^2 - \big(c- \frac{1}{2} \big)^2 - \big(d- \frac{1}{2} \big)^2 \geq 0$ which is possible only when R.H.S is zero i.e., $a=b=c=d= \frac{1}{2}$, since the R.H.S is always non-positive.