Cubic conundrum 2

Let the three positive real roots be $a$ , $b$ , and $c$ . By Vieta's formula the monic cubic polynomial is

$\begin{aligned} f(x) & = x^3 -(a+b+c)x^2 + (ab+bc+ca)x - \blue{64} & \small \blue{\text{where } abc = 64} \\ \implies f(-1) & = -1 - \blue{(a+b+c)} - \red{(ab+bc+ca)} - 64 & \small \blue{\text{By AM-GM inequality, }a+b+c \ge 3\sqrt{abc} = 3\cdot 4} \\ & \le -1 - \blue{12} - \red{48} - 64 & \small \red{\text{and }ab+bc+ca \ge 3\sqrt{(abc)^2} = 3\cdot 4^2} \\ & = \boxed{-125} & \small \blue{\text{Equality occurs when }a=b=c=4.} \end{aligned}$

Reference: AM-GM inequality

Let the polynomial be $f(x)=(x-u)(x-v)(x-w)$ where $u,v,w$ are non-negative real numbers.

Then $f(0)=-uvw$ so that $uvw=64$ and $f(-1)=-(1+u)(1+v)(1+w)$ .

By symmetry, we want $u=v=w=4$ , giving $f(-1)=\boxed{-125}$ .

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To check, though, we can maximise this with a Lagrangian multiplier. Define $F(u,v,w,L)=-(1+u)(1+v)(1+w)+L(uvw-64)$

Setting the partial derivatives to zero, this gives $\begin{aligned} Lvw &=(1+v)(1+w) \\ Lwu&=(1+w)(1+u) \\ Luv &=(1+u)(1+v) \\ uvw&=64 \end{aligned}$

We can solve these by multiplying the first three equations together and substituting in for $uvw$ ; the algebra gets a little messy but in the end we find $u=v=w=\frac{1}{\sqrt{L}-1}$

so that, again, $u=v=w=4$ .