Calculus or Algebra?

$\huge\color{#302B94}{\boxed{ \color{#007fff}{\mathcal{Algebraic ~method}}}}$ $xyz=3 \Rightarrow \color{#D61F06}{yz=\dfrac{3}{x}}....(1)$ $xy+zx=-5-yz= -5-\dfrac{3}{x}..(\text{Using 1})$ $\Rightarrow \color{#D61F06}{y+z=\dfrac{-5}{x} - \dfrac{3}{x^2}}....(2)$ Now from (1) and (2) and using $\color{#D61F06}{(y-z)^2\geq0~ or ~(y+z)^2\geq 4yz}$ , we get $\dfrac {(5x+3)^2}{x^4}\geq \dfrac{12}{x}$ $\Rightarrow 12x^3-25x^2-30x-9 \leq 0$ $\Rightarrow (x-3)(12x^2+11x+3)\leq 0$ Since $12x^2+11x+3>0$ for all real x we can safely right $\large x\leq3$ Hence maximum value of x=3 and $\large 2(3)^3=54$

This proof is more intuitive than it is rigorous or elegant. However, this doesn't make it any less valid!

From $xyz=3$ and $xy+yz+xz=-5$ , we can conclude that only one of $x,y$ or $z$ is positive whereas another two are negative. Without loss of generality, let $x>y \geq z$ (ie. $x$ is positive, $y$ and $z$ are negative.)

We obtain the following equations:

$x=\frac{3}{yz}$

$x=\frac{-5-yz}{y+z}$

Equating both equations yield ${(yz)}^{2}+5yz+3(y+z)=0$ . Viewing this as a quadratic equation in $yz$ , we obtain $yz=\frac{-5\pm \sqrt{25-12(y+z)}}{2}$ . Since $yz$ is positive, the we replace the $\pm$ with $+$ .

Therefore, $yz=\frac{-5 + \sqrt{25-12(y+z)}}{2}$ . We see that if $(y+z)$ is maximized within the allowed constraints, then $yz$ is minimized within the allowed constraints. Looking at $x=\frac{3}{yz}$ again, to maximize $x$ , minimizing $yz$ within the allowed constraints is exactly what we want to do!

Therefore, it is natural for us to seek ways to maximize $y+z$ within the allowed constraints. If we succeed in doing so, $yz$ will be minimized, which is what we want.

Note that minimizing $-(y+z)$ is equivalent to maximizing $(y+z)$ . Since $(y+z)<0$ , to maximize it, we are bringing it as close to $0$ as possible. This is similar with minimizing $-(y+z)>0$ .

By AM-GM (note that $-z \geq -y >0)$ , we have

$-y-z$

$\geq 2\sqrt{(-y)(-z)}$

$=2\sqrt{yz}$

Therefore, $\min_{-y-z}=2\sqrt{yz}$ .

However, since $-y-z \geq -2y$ , we can also say that $\min_{-y-z}=-2y$ .

Equating the both gives $2\sqrt{yz}=-2y$ , which implies $y=z$ when the expression $-(y+z)$ attains its minimum.

Substitute $y=z$ back into our original equations gives $xy^{2}=3$ and $2xy+y^{2}=-5$ .

From the first equation, we get $y=-\sqrt{\frac{3}{x}}$ , $y^{2}=\frac{3}{x}$ . Substitute this into the second equation gives the following cubic equation in terms of $x$ :

$12x^{3}-25x^{2}-30x-9=0$

$(x-3)(12x^{2}+11x+3)=0$ .

The only real solution is $x=3$ . We check that $x=3,y=z=-1$ indeed satisfies the two equations above, hence the solution.

Let $x, y$ and $z$ be the roots of a cubic polynomial function $P(m)$ . Then, by Vieta's formulas, $P(m) = m^3+km^2+(xy+yz+xz)m-xyz$ , which gives us $P(m) = m^3+km^2-5m-3$ for some $k$ .

Next, observe the graph of $P(m)$ . We notice that the graph rises to the right and falls to the left, meaning its local maxima is to the left of its local minima. Also, as with all cubic polynomial functions, $P(m)$ has three real roots if and only if its local maxima is greater than or equal to zero, and when its local minima is less than or equal to zero.

Now, obviously, $x, y$ and $z$ must be real numbers. Since all real numbers become nonnegative when squared, it is implied that in polynomial $P(m) = m^3+km^2-5m-3$ , $m^2$ is always positive, and so, this means that for all $m$ except for 0, the value of $P(m)$ increases as k increases, and decreases as k decreases, as shown in the figure below.

This also implies that as k decreases, the value of the local maxima also decreases, but, the value of the third root also increases, as shown in the illustration below.

So, in short, we have to find the minimum value of k for which the polynomial $P(m)$ has local maxima exactly equal to zero. But, when that happens, notice that two of the roots of $P(m)$ will be equal, as the maxima becomes tangent to the x-axis. Since the roots of $P(m)$ are $x, y$ and $z$ , this implies that $y=z$ to get the maximum value of x. Substituting,

$2xy+y^2 = -5$

$xy^2=3$

Then, solving the systems, we get $x=3$ , and $2x^3 = \boxed {54}$