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

[anemone]

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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[anemone]

No one answered last week's POTW. (Sadface) However, you can find the suggested solution below:

Spoiler

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 $\dfrac{1}{z}=\dfrac{-1+\sqrt{3}}{3}$.