MHB Finding Extremes of x^3+y^3+z^3+xyz for Real Numbers with Constraints

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The problem involves finding the maximum and minimum values of the expression x^3 + y^3 + z^3 + xyz under the constraints x + y + z = 4 and 1/x + 1/y + 1/z = 1/3. The discussion highlights the lack of responses to the previous problem of the week (POTW), indicating a potential challenge in engaging participants. A suggested solution is provided, although no specific answers or methods have been detailed in the discussion. The focus remains on solving the mathematical expression within the given constraints. Engaging with this problem could enhance understanding of algebraic expressions and optimization techniques.
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Here is this week's POTW:

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Real numbers $x,\,y$ and $z$ satisfy $x+y+z=4$ and $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{3}$. Find the largest and smallest possible value of the expression $x^3+y^3+z^3+xyz$.

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No one answered last week's POTW. (Sadface) However, you can find the suggested solution below:
Note that $(x+y+z)^3=x^3+y^3+z^3+3(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2)+6xyz$ while $3(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)xyz=3(x+y+z)(xy+xz+yz)=3(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2)+9xyz$. Thus,
$(x+y+z)^3-3(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)xyz=x^3+y^3+z^3-3xyz$.

By assumptions, $x+y+z-4$ and $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{3}$. Hence $63-4xyz=x^3+y^3+z^3-3xyz$, implying $x^3+y^3+z^3+xyz=64$.

Consequently, the expression $x^3+y^3+z^3+xyz$ has only one value 64.

An example of numbers satisfying the conditions is $x=1,\,y=\dfrac{3-3\sqrt{3}}{2}$ and $z=\dfrac{3+3\sqrt{3}}{2}$. Then $\dfrac{1}{x}=1,\,\dfrac{1}{y}=\dfrac{-1-\sqrt{3}}{3}$ and $\dfrac{1}{z}=\dfrac{-1+\sqrt{3}}{3}$.
 
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