Keep it real (and distinct)

If $f(x)$ has four distinct real roots, then $f'(x)$ will have three (as in Pi Han Goh's "Method 2") and $f''(x)=12x^2+24x+2b$ will have two, by the mean value theorem (or Rolle's theorem). The discriminant of $f''(x)$ , namely, $96(6-b)$ , will be positive, so $b<6$ .

It is not hard to see that all $b<6$ are attained. Note that $(x+1)^4=x^4+4x^3+6x^2+4x+1$ . We can perturb the coefficients a bit (as in Mark Hennings' solution) to create four distinct real roots while keeping the coefficient of $x^2$ as close to $6$ as we wish. Consider $\left((x+1)^2-\epsilon\right)\left((x+1)^2-2\epsilon\right)=(x+1)^4-3\epsilon(x+1)^2+2\epsilon^2$ for positive $\epsilon$ . The coefficient of $x^2$ will be $6-3\epsilon$ , and we have four distinct real roots, $-1\pm \sqrt{\epsilon},-1\pm \sqrt{2\epsilon}$ . Thus the supremum of $B$ is $\boxed{6}$ .

DISCLAIMER: All the methods below are wrong because I have only proven that $B<6$ , but I didn't prove that $B$ has a supremum of 6.

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Method 1: Let $p>q>r>s$ be the real roots to the polynomial $x^4 + 4x^3 + bx^2 + cx + d$ . Then by Vieta's Formula , the following equations are true, $\begin{cases} p + q + r + s = -4 \\ pq + pr + ps + qr + qs + rs = b. \end{cases}$ Since $p>q>r>s$ is true, the following inequality holds, $(p-q)^2 + (p-r)^2 + (p-s)^2 + (q-r)^2 + (q-s)^2 + (r-s)^2 > 0$ Expanding and simplifying it gives $3(p^2 + q^2 + r^2 + s^2) - 2(pq + pr + ps + qr + qs + rs) > 0 .$ With $p^2 + q^2 + r^2 + s^2 = (p+q+r+s)^2 - 2(pq + pr + ps + qr + qs + rs)$ , we can solve for the range of $b$ , $3( (-4)^2 - 2b) - 2b > 0 \qquad \Rightarrow \qquad b < \boxed{6} .$

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Method 2: Since the quartic polynomial $f(x):=x^4 + 4x^3 + bx^2 + cx + d$ has 4 distinct real roots, then the cubic polynomial $f'(x) = 4x^3 + 12x^2 + 2bx + c$ has 3 distinct real roots. So the (cubic) discriminant of $f'(x)$ is positive, $(12)^2 (2b)^2 - 4(4)(2b)^3 - 4(12)^3 (c) - 27(4)^2 (c)^2 + 18(4)(12)(2b)(c) > 0 \quad \Rightarrow \quad c^2 (27) + c(432 - 108b) + (8b^3 - 36b^2 ) < 0.$ This quadratic inequality (in $c$ ) has a positive (quadratic) discriminant , $(432 - 108b)^2 - 4(27)(8b^3 - 36b^2) < 0\quad \Rightarrow \quad b< \boxed6.$

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Method 3: Since the quartic polynomial $f(x):=x^4 + 4x^3 + bx^2 + cx + d$ has 4 distinct real roots, then there exists a straight line $\ell(x) : mx + n$ that is (below and) tangent to $f(x)$ at exactly 2 points. Thus, $f(x) - \ell(x)$ has double roots, so for reals $j\ne k$ , $\begin{aligned} f(x) - \ell(x) &=& (x-j)^2 (x-k)^2 \\ (x^4 + 4x^3 + bx^2 + cx + d) - (mx + n) &=& (x-j)^2 (x-k)^2 \\ x^4 + x^3(4) + x^2 (b) + x(c-m) + (d-n) &=& x^4 + x^3(-2j - 2k) + x^2(j^2 + 4jk + k^2 )+ (-2j^2k - 2jk^2) + (j^2 k^2) \end{aligned}$ By comparing the coefficients of the powers of $x$ , we get $\begin{cases} -2j - 2k = 4 \\ j^2 + 4jk + k^2 = b \end{cases} \quad \Rightarrow \quad \begin{cases} j + k = -2 \\ jk = \frac{b-4}2. \end{cases}$ And finally $(j-k)^2 > 0 \quad \Rightarrow \quad j^2 + k^2 - 2jk > 0\quad \Rightarrow \quad (j^2 + k^2 + 4jk) - 6jk > 0 \quad \Rightarrow \quad b - 6 \left( \frac{b-4}2\right) > 0 \quad \Rightarrow \quad b < \boxed6.$

If $f(x) = x^4 + 4x^3 + bx^2 + cx + d$ has four distinct real zeros, then $f(x-1) \; = \; x^4 + (b-6)x^2 + (c-2b+8)x + (d-c+b-3)$ has four distinct real zeros that add to zero. If these zeros are $\alpha,\beta,\gamma,\delta$ , then $b-6 \; = \; \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta \; = \; \tfrac12\big[(\alpha+\beta+\gamma+\delta)^2 - (\alpha^2 + \beta^2 + \gamma^2 +\delta^2)\big] \; = \; -\tfrac12(\alpha^2 + \beta^2 + \gamma^2 + \delta^2)$ so that $b \; = \; 6 - \tfrac12(\alpha^2+\beta^2+\gamma^2+\delta^2) \; < \; 6$ By choosing $f(x) \; = \; \big((x+1)^2-u^2\big)\big((x+1)^2 - v^2\big) \; = \; (x+1)^4 - (u^2+v^2)(x+1)^2 + u^2v^2$ for distinct positive reals $u,v$ , we see that we can certainly achieve $b = 6 - u^2 - v^2$ for any such $u,v$ . Thus $B = (-\infty,6)$ , so the supremum is $\boxed{6}$ .