If $x + y + z = 2$, then show $\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$, added a second question(Problem 2).

Question

Problem number 1:

The problem is that $x, y, z$ are proper fractions, and each one of them is greater than zero.

Given $x + y + z = 2$, prove $$\frac{x y z}{(1-x)(1-y)(1-z)} \geq 8$$

I have tried to solve this using AM $\geq$ GM inequality.

Attempt : $$\frac{\dfrac{1-x}{x} + \dfrac{1-y}{y} + \dfrac{1-z}{z}}{3}\geq\left(\frac{(1-x)(1-y)(1-z)}{xyz}\right)^{1/3}$$

What should I do to calculate the value of $\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$?

Doubt:

• Let $a, b, c, d$ all be positive numbers.
• Let $a > c$ and $b > d$.
• Can we say that $\dfrac {a}{b} > \dfrac{c}{d}$?

Problem number 2:

$x,y,z$ are unequal positive quantities, prove that $(1+x^3)(1+y^3)(1+z^3) > (1+xyz)^3$

My attempt:

$ \frac{x^3+y^3+z^3}{3} > xyz$

$ => (1+x^3)+(1+y^3)+(1+z^3) > 3(1+xyz)$

now by cubing the both sides we get,

$ => \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > (1+xyz)^3$ ---(eqn. 1)

$ \frac{(1+x^3)+(1+y^3)+(1+z^3)}{3} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}^{\frac{1}{3}}$

$=> \frac{(1+x^3)+(1+y^3)+(1+z^3)}{27} > {\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)}$ ---(eqn. 2)

So, now if we do : $\frac{(eqn. 1)}{(eqn. 2)}$

then the result will be :

${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} > (1+xyz)^3$

and this is the correct result for all $x,y,z$ greater than zero.

But if we do $\frac{(eqn. 2)}{(eqn. 1)}$ then the result will be :

${\biggl((1+x^3)(1+y^3)+(1+z^3)\biggl)} < (1+xyz)^3$

So, my question is, how to prove it correctly without doing the division operation, cause when we divide the two equations then we get two separate results.

Answer 1

The first inequality:

We need to prove that: $$\frac{8xyz}{\prod\limits_{cyc}(x+y+z-2x)}\geq8$$ or $$xyz\geq\prod_{cyc}(y+z-x)$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0,$$ which is Schur.

The second inequality it's just Holder for three sequences: $$\prod_{cyc}(1+x^3)\geq\left(\sqrt[3]{1\cdot1\cdot1}+\sqrt[3]{x^3y^3z^3}\right)^3=(1+xyz)^3.$$

Answer 2

We have $x+y+z=2$ so $1-x+1-y+1-z=3-(x+y+z)=1$ using Am-GM inequality twice once for $$x+y+z\geq 3(xyz)^{\frac{1}{3}}$$ and second for $$1-x+1-y+1-z\geq 3((1-x)(1-y)(1-z))^{\frac{1}{3}}$$,cubing both inequalities and dividing the two should yield the result.

Answer 3

$\dfrac ab-\dfrac cd=\dfrac{ad-bc}{be}$ will be $\ge0$

only if $ad-bc>0\iff \dfrac ab>\dfrac cd$

Alternatively

Let $a=c+p,b=d+q,p,q>0$

$\dfrac ab-\dfrac cd=\dfrac{c+p}{d+q}-\dfrac cd=\dfrac{d(c+p)-c(d+q)}{...}$

$=\dfrac{dp-cq}{...}$

will be $>0$

only if $dp-cq>0\iff \dfrac dc>\dfrac qp$

as $c,d,p,q>0$