Computing the composition $f(f(x)) = 0$ yields: $f(f(x)) = (x^2 + ax + b)^2 + a(x^2 + ax + b) + b = 0$; or $x^4 + 2ax^3 + (a^2 + 2b + a)x^2 + (2ab + a^2)x + (b^2 + ab + b) = 0$ (i) If (i) has 4 distinct real roots, then $(x - p)(x - q)(x - r)(x - s) = 0$ and by Vieta's Formulae: $p + q + r + s = -2a$; $pq + pr + ps + qr + qs + rs = a^2 +2b + a$; $pqr + pqs + prs + qrs = -2ab - a^2$; $pqrs = b^2 + ab + b$. WLOG, let $p + q = -1$ so that we next obtain: $r + s = 1 - 2a$; $pq + rs - r - s = a^2 + 2b + a$; $pq(r + s) - rs = -2ab - a^2$; $pqrs = b^2 + ab + b$ and if we substitute the 1st equation into the 2nd and 3rd equations above, then we have: $pq + rs = a^2 + 2b - a + 1$ (ii); $(1 - 2a)pq - rs = -2ab - a^2$ (iii) which solves as: $pq = b + 1/2, rs = a^2 + b - a + 1/2$ (iv). We can now substitute (iv) into $pqrs = b^2 + ab + b$ to produce: $(b + 1/2)(a^2 + b - a + 1/2) = b^2 + ab + b$; or $b(a^2 - 2a) = -(a^2)/2 + a/2 - 1/4$; or $b = g(a) = (-2a^2 + 2a - 1)/[4(a^2 - 2a)]$ (v). Taking the derivative of $g(a)$ equal to zero yields the critical points: $g'(a) = (a^2 + a - 1)/[2(a^2 - 2a)^2] = 0 \Rightarrow a = \frac{-1 \pm \sqrt{5}}{2}$; and the 2nd derivative of $g(a)$ at these critical points yields:  $g''( (-1+\sqrt{5})/2 ) = 3/4 + 7/(4\sqrt{5}) > 0$ (a minimum); $g''( (-1 - \sqrt{5})/2 ) = 3/4 - 7/(4\sqrt{5}) < 0$ (a maximum) Thus, the maximum value of $b$ computes to: $b = g( (-1 - \sqrt{5})/2 ) = -[1 + \sqrt{5}]/8 = (-1/4)[1 + \sqrt{5}]/2 \le -1/4$ QED