Can Derivative Work?

My solution bases on Jerry Han Jia Tao 's one, but simpler.

From the system we have $x^2+y^2+(4-x-y)^2=6 \Leftrightarrow 2x^2+2y^2+2xy-8x-8y+10=0$ .

Rewrite in terms of $y$ :

$2y^2+2(x-4)y+(2x^2-8x+10)=0$ (1) .

$\Delta'=(x-4)^2-2(2x^2-8x+10)=-3x^2+8x-4$ .

Equation (1) has real solutions if and only if $\dfrac 23 \leq x \leq 2$ .

But when we solve (1) , we have $y=\dfrac{4-x\pm\sqrt{-3x^2+8x-4}}{2}$ , which is annoying.

So here's what we'll do next. By this method we could generalize the result easily.

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Rewrite the system in terms of $y,z$ :

$\begin{cases} y+z=4-x \\ y^2+z^2=6-x^2 \end{cases} \Leftrightarrow \begin{cases} y+z=4-x \\ yz=x^2-4x+5 \end{cases}$

We calculate $y+z$ and $yz$ because all symmetric expressions in terms of $y$ and $z$ can be expressed as an expression in terms of $y+z$ and $yz$ . By this we can avoid roots and calculating mistakes.

For example, $y^3+z^3=(y+z)^3-3yz(y+z)=(4-x)^3-3(4-x)(x^2-4x+5)=2x^3-12x^2+15x+4$ .

Hence, $P=x^3+y^3+z^3=3x^3-12x^2+15x+4$ .

The last part is to find the local maxima and local minima of this expression, using Derivative.

$P'=9x^2-24x+15$

$P'(1)=P'\left(\dfrac53\right)=0$

Thus, the result $=\left|P(1)-P\left(\dfrac53\right)\right|=\dfrac49$ , and $4+9=\boxed{13}$ .

For the sake of algebra:

$x+y+z=m$ $x^2+y^2+z^2=n$ $xy+yz+xz=\frac{m^2-n}{2}$

Let $b=x+y+z=m$ , $c=xy+yz+xz=\frac{m^2-n}{2}$ and $d=xyz$ .

$x^3+y^3+z^3=3xyz+(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=3d+b(n-c)$

Note that this expression only has $d$ as a variable, and $b$ and $c$ are constants given values of $m$ and $n$ .

Hence $P_{max}-P_{min}=3(d_{max}-d_{min})$

Consider the cubic polynomial $r^3-br^2+cr-d$

$x,y$ and $z$ are real, so its discriminant is nonnegative.

$\Delta_1 = b^2c^2-4c^3-4b^3d-27d^2+18bcd \geq 0$

$27d^2+(4b^3-18bc)d+4c^3-b^2c^2 \leq 0$ $\frac{18bc-4b^3 - \sqrt{16b^6-144b^4c+324b^2c^2-432c^3+108b^2c^2}}{54} \leq d \leq \frac{18bc-4b^3 + \sqrt{16b^6-144b^4c+324b^2c^2-432c^3+108b^2c^2}}{54}$ $d_{max}-d_{min}=\frac{2\sqrt{16b^6-144b^4c+432b^2c^2-432c^3}}{54}$ $P_{max}-P_{min}=3(d_{max}-d_{min})=\frac49 \sqrt{b^6-9b^4c+27b^2c^2-27c^3}$ $=\frac49 \sqrt{(b^2-3c)^3}$

Substituting $b=m$ and $c=\frac{m^2-n}{2}$ , we get the final answer of $\boxed{\dfrac49 \sqrt{(\dfrac{3n-m^2}{2})^3}}$

We use Newton's sums to obtain the ff. equations:

$xy+yz+xz = \frac{(x+y+z)^2-(x^2+y^2+z^2)}{2} = \frac{4^2-6}{2} = 5$ .

$x^{3} + y^{3} + z^{3} = (x+y+z)(x^{2}+y^{2}+z^{2}-(xy+yz+zx)) +3xyz$

So, $x^{3} + y^{3} + z^{3}$ will be maximized if $xyz$ is maximized and minimized if $xyz$ is minimized.

Letting $x,y,z$ be the roots of a cubic equation $n^3-(x+y+z)n^2+(xy+yz+xz)n-xyz = n^3-4n^2+5n-(xyz)$ ,

We see that the graph of $f(n)$ will be either transformed upwards or downwards as $xyz$ increases or decreases respectively. But, in order to have real roots, we must have the x-axis between the graph's local maxima and minima, or exactly on one of them. To maximize $xyz$ , we must have the maxima on the x-axis, and to minimize it, we must have the minima on the x-axis. Either case, the graph is tangent to the x-axis, implying a double root. Having a double root implies WLOG $x=z$ . Therefore, we just substitute this into the equations to obtain $2x+y=4$ and $2x^2+y^2 = 6$ . Solving gives us either $x=z=1, y=2$ or $x=z=\frac{5}{3}, y = \frac{2}{3}$ . This gives us values of $x^3+y^3+z^3$ as $10, \frac{86}{9}$ . So, $10$ is the maxima and $\frac{86}{9}$ is the minima. Their difference is $\frac{4}{9}$ , so our answer is $4+9 = \boxed{13}$

$xy + yz + zx = \dfrac{(x+y+z)^{2}-(x^{2}+y^{2}+z^{2}}{2} = \dfrac{4^2 -6}{2} = 5$
$x^{3} + y^{3} + z^{3} - 3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-(xy+yz+zx))$
Let $x^{3} + y^{3} + z^{3} = P$
$xyz = \dfrac{p-4}{3}$
$x,y,z$ are the roots of the equation,
$f(a) = a^{3} - (x+y+z)a^{2} + (x^{2}+b^{2}+z^{2})a - (xyz) = a^{3} - 4a^{2} + 5a -\left(\dfrac{p-4}{3} \right)$
$f'(a) = 3a^2 - 8a + 5 = (3x-5)(x-1)$
The two roots of $f'(a)$ are $a = 1, a = \dfrac{5}{3}$
$f''(a) = 6a-8$
$f''(1) < 0 \to a = 1$ is a local maxima.
$f''\left(\dfrac{5}{3}\right) > 0 \to a = \dfrac{5}{3}$ is a local minima.
$\displaystyle \lim_{a \to -\infty} f(a)= -\infty , \lim_{a \to \infty}f(a) = \infty$
For the equation to have 3 real roots, the maxima must lie above or on the x-axis and the minima must lie below or on the x-axis .
$\therefore f(1) \geq 0$
This gives us,
$p \leq 10$
$\therefore f\left(\dfrac{5}{3}\right) \leq 0$
This gives us,
$p \geq \dfrac{86}{9}$
$\dfrac{86}{9} \leq x^{3} + y^{3} + z^{3} \leq 10$
$p_{max} - p_{min} = 10 - \dfrac{86}{9} = \dfrac{4}{9}$
$a + b = 4 + 9 = 13$

$P_{max}$ occurs when $x,y,z$ are a permutation of roots of,
$f(a) = a^3 - 4a^2 + 5a - 2 = (a-1)^{3}(a-2) = 0 \to (x,y,z) \in (1,1,2)$
$P_{min}$ occurs when $x,y,z$ are a permutation of roots of,
$f(a) = a^3 - 4a^2 + 5a - \dfrac{50}{27} = (a-\dfrac{5}{3})^{2}(a-\dfrac{2}{3}) = 0 \to (x,y,z) \in ( \dfrac{5}{3}, \dfrac{5}{3}, \dfrac{2}{3})$