Is this a Quadratic? Wait it is not

We have the quadratic equation $ax^2 + bx + c$ that must be non-negative for all $x \in \mathbb{R}.$ Two necessary & sufficient conditions must exist here:

1) $a > 0$ ;

2) By the Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ we need the discriminant value to satisfy $b^2 - 4ac \le 0$ .

Now, let $f(a,b,c) = 4a^3 - b^3 + 4c^3$ and $g(a,b,c) = b^2 - 4ac \le 0$ be two functions in 3-variables. We can apply Lagrange Multipliers to compute the minimum value of $f$ : $grad(f) = \lambda \cdot grad(g)$ .

This yields partial derivatives:

$12a^2 = \lambda (-4c) (i);$ $-3b^2 = \lambda (2b) (ii);$ $12c^2 = \lambda (-4a) (iii).$

Setting $(i)$ equal to $(iii)$ produces $a^3 = c^3 \Rightarrow a^3 - c^3 = (a-c)(a^2 + ac + c^2) = 0 \Rightarrow c = a, \frac{-1 \pm \sqrt{3}i}{2} \cdot a$ . The former value is only allowed since $a, c \in \mathbb{R}$ . Setting $(i)$ equal to $(ii)$ yields $b = 2a$ , and we now substitute these two (2) above values into $g(a,b,c)$ to compute:

$b^2 - 4ac \le 0 \Rightarrow (2a)^2 - 4a^2 \le 0 \Rightarrow 4a^2 - 4a^2 \le 0 \Rightarrow 0 \le 0 (iv).$

It turns out that $(iv)$ is independent of $a$ , which yields the result $(a,b,c) = (t, 2t, t)$ (for $t > 0$ ). A final substitution of this critical point into $f$ gives us: $f(t,2t,t) = 4t^3 - (2t)^3 + 4t^3 = 8t^3 - 8t^3 = \boxed{0}.$

Well, then I am writing an algebraic solution.

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Note that since $ax^2+bx+c \geq 0$ , we get $b^2 \leq 4ac$ and that $a \geq 0$ and $c \geq 0$ .

Now since we need to minimize $4a^3-b^3+4c^3$ , this can be done by minimising $4(a^3+c^3)$ and maximising $b^3$ which means maximizing $b$ .

As $b^2 \leq 4ac$ , maximum of $b$ will occur when $b=2\sqrt{ac}$ implying $b^3=8\left(ac\right)^{\dfrac{3}{2}}$ .

And by using AM-GM Inequality (since both $a$ and $c$ are positive) we get $4(a^3+c^3) \geq 8(ac)^{\dfrac{3}{2}}$ .

Hence, minimum value is $8(ac)^{\dfrac{3}{2}}-8(ac)^{\dfrac{3}{2}}=0$ .

Equality occurs when $a=c=t , b=2t$ for any $t \geq 0$ .