Finding Both Minimum And Maximum

Let a function $f\left( p \right) \, =\, { p }^{ 3 }\, +\, A{ p }^{ 2 }\, +\, Bp\, +\, C$ have roots $x$ , $y$ , and $z$ .

By Vieta's Formulas , we know that

$-A\, =\, x\, +\, y\, +\, z\, =\, 4\, \Rightarrow \, A\, =\, -4$

and

${ \left( x\, +\, y\, +\, z \right) }^{ 2 }\, =\, { 4 }^{ 2 }\\ { x }^{ 2 }\, +\, { y }^{ 2 }\, +\, { z }^{ 2 }\, +\, 2\left( xy\, +\, xz\, +\, yz \right) \, =\, 16\\ 6\, +\, 2B\, =\, 16\, \Rightarrow \, B\, =\, 5$

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${ x }^{ 6 }\, +\, { y }^{ 6 }\, +\, { z }^{ 6 }$ will have extrema when $f\left( p \right)$ has a multiple root. This will occur when $f\left( p \right)$ has a critical point with an ordinate of 0.

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$f^{ ' }\left( p \right) \, =\, 3{ p }^{ 2 }\, -\, 8p\, +\, 5\, =\, 0\\ p\, =\, 1\, and\, \frac { 5 }{ 3 }$

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Since those are double roots, the other roots must be $2$ and $\frac { 2 }{ 3 }$ respectively.

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$max\left( { x }^{ 6 }\, +\, { y }^{ 6 }\, +\, { z }^{ 6 } \right) \, =\, { 1 }^{ 6 }\, +\, { 1 }^{ 6 }\, +\, { 2 }^{ 6 }\, =\, 66\\ min\left( { x }^{ 6 }\, +\, { y }^{ 6 }\, +\, { z }^{ 6 } \right) \, =\, { \left( \frac { 5 }{ 3 } \right) }^{ 6 }\, +\, { \left( \frac { 5 }{ 3 } \right) }^{ 6 }\, +\, { \left( \frac { 2 }{ 3 } \right) }^{ 6 }\, =\, 42\frac { 232 }{ 243 } \\ 66\, +\, 42\frac { 232 }{ 243 } \, =\, \boxed { 108\frac { 232 }{ 243 } \, \approx \, 109 }$

Relevant wiki: Classical Inequalities - Problem Solving - Advanced

Because $(x+y+z)^2 = x^2+y^2 + z^2 + 2(xy+xz+yz)$ , then $xy+xz+yz = 5$ .

Let $f(X)$ be a monic cubic polynomial with roots $x,y$ and $z$ . Then, by Vieta's formula , $f(X) = X^3 - 4X + 5X + D$ , where $D =-xyz$ .

Since $x,y$ and $z$ are real numbers, then $f(X)$ has all real roots, so its cubic discriminant is non-negative:

$b^2 c^2 - 4ac^3 - 4b^3 d - 27a^2 d^2 + 18abcd \geq 0 \; ,$

where $a = 1, b = -4, c = 5, d = D$ . Substituting these values and solving this inequality yields $-2 \leq D \leq -\dfrac{50}{27}$ .

Let $P_n = x^n +y^n + z^n$ , then $P_1 = 4, P_2=6$ and we want to find $\max(P_6) + \min(P_6)$ .

Using the algebraic identity, $x^3+y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2 + z^2 - xy -xz - yz)$ , we get $P_3 = -3D + 4$ .

By Newton's sum , we have

$\begin{aligned} && P_4 - 4P_3 + 5P_2 + DP_1 = 0 \Rightarrow P_4 = -16D - 14 \\ && P_5 - 4P_4 + 5P_3 + DP_2 = 0 \Rightarrow P_5 = -55D - 76 \\ && P_6 - 4P_5 + 5P_4 + DP_3 = 0 \Rightarrow P_6 = 3(D^2 - 48D - 78) = 3( (D-24)^2 -654 ) \; . \end{aligned}$

Since $g(D) = 3( (D-24)^2 -654 )$ is a strictly decreasing function in the interval $-2 \leq D \leq -\dfrac{50}{27}$ , then $\max(P_6) = \max(g(D)) = g(-2) = 66$ and $\min(P_6) = \min(g(D)) = g\left( -\dfrac{50}{27} \right) =42 \dfrac{232}{243}$ .

Hence, our answer is $\left \lfloor 66 + 42 \dfrac{232}{243} \right \rceil = \boxed{109}$ .