If $x + y + z = 1$, what is $\min(xyz)$ where $x$, $y$ and $z$ are positive numbers?

Question

$x,y,z$ are positive numbers and given $x + y + z = 1$. I was initially required to prove that $xy+yz+xz - 3xyz \leq \frac{1}{4}$. I then manipulated such that I was required to prove that $xyz \geq \frac{1}{36}$ (using $x^3 +y^3 +z^2 -3xyz = (x+y+z)(x^2+y^2+z^-xy-yz-xz)$ I know how to prove that the upper limit of $xyz$ is $\frac{1}{27}$, but I am not too sure how to find the lower limit of $xyz$. Any help would be appreciated, and if you have any other suggestions on how to solve the original inequality - that would be appreciated as well. Apologies if I am not correctly following any of the posting guidelines as this is my first question.

Answer 1

I think $\min(xyz)$ is $0$, when any of x,y,z is reaching $0$.

Answer 2

As it was already pointed out, there is no minimum. Since $x,y,z>0$, we have that $xyz > 0$. But if you fix any $0 < \varepsilon < \frac 14$ and consider the point $(\sqrt\varepsilon, \sqrt\varepsilon, 1-2\sqrt\varepsilon)$, you see that the objective function value is given by $$ \sqrt \varepsilon \sqrt \varepsilon (1-2\sqrt \varepsilon) = \varepsilon (1-2 \sqrt \varepsilon), $$

which is bounded by $\varepsilon$. So, you can find points $(x,y,z)$ in the admissible region such that $xyz < \varepsilon < \frac 14$. Since $xyz>0$, we conclude that $0$ is the infimum but not the minimum.

Answer 3

The inequality, which you "was initially required to prove", we can prove by the following way.

We need to prove that: $$xy+xz+yz-3xyz\leq\frac{1}{4}$$ or $$(x+y+z)^3\geq4(x+y+z)(xy+xz+yz)-12xyz$$ or $$\sum_{cyc}(x^3+3x^2y+3x^2z+2xyz)\geq\sum_{cyc}(4x^2y+4x^2z)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+2xyz)\geq0,$$ which is true because by Schur$$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0$$ and $$\sum_{cyc}xyz\geq0.$$