If $x,y,z>0$ and $x+y+z=1$ then $\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$

Question

If $x,y,z>0$ and $x+y+z=1$,
then: $$\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$$

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$$\frac{xyz}{(1-x)(1-y)(1-z)}=\frac{x}{(1-x)}\frac{y}{(1-y)}\frac{z}{(1-z)}$$

Let $\frac{x}{(1-x)}=a$, $\frac{y}{(1-y)}=b$, $\frac{z}{(1-z)}=c$

I am stuck here.

Answer 1

You have: $$x+y+z=1$$ So, write: $$1-x=y+z$$ Using inequality $AP\geq GP$, $$y+z\geq 2\sqrt{yz}$$ So, $$(1-x)\geq2 \sqrt{yz}$$ Similarly, $$(1-y)\geq2 \sqrt{xz} \hspace{0.5cm} {and} \hspace{0.5cm} (1-z)\geq2 \sqrt{xy}$$ Multiplying all: $$(1-x)\cdot (1-y) \cdot (1-z) \geq 8 \cdot xyz$$

Divide:$$\frac{(1-x)\cdot (1-y)\cdot (1-z)}{xyz}\geq 8$$ Or, $$\frac{xyz}{(1-x)(1-y)(1-z)}\leq \frac 1 8$$ In-equality will be reversed on dividing (why?)

Answer 2

Write $8xyz\le (x+y)(y+z)(x+z)$ and use arithmetic geometric inequality.

Answer 3

Hint.- Just to give another way. Developing the product and simplify you have the equivalent inequality $$7xyz\le xy+xz+yz$$ and you have too $$\frac13\ge\sqrt[3]{xyz}$$

Answer 4

When you can separate variables in a simple way a good try is by convexity/concavity. In this case you can take the logarithm and use Jensen on the function $$\log\left(\frac{x} {1-x}\right)$$ which you can prove to be concave by taking the second derivative.