Minimize Samuel's fraction

When $x, y, z, t$ all are negative real numbers (say $x=-p, y=-q, z=-r$ and $t=-s$ respectively), then the expression becomes $|(\frac{p^2-p+1}{p})(\frac{q^2-q+1}{q})(\frac{r^2-r+1}{r})(\frac{s^2-s+1}{s})| = |(p+\frac{1}{p}-1)(q+\frac{1}{q}-1)(r+\frac{1}{r}-1)(s+\frac{1}{s}-1)| \ge |(2-1)(2-1)(2-1)(2-1)| = 1$ as $k+\frac{1}{k} \ge 2$ for positive real number $k$ . So minimum value is 1.

We can rewrite the given equations as:

$\large [x+1+\frac{1}{x}][y+1+\frac{1}{y}][z+1+\frac{1}{z}][t+1+\frac{1}{t}]$

$\large [x+1+\frac{1}{x}] \geq 3$ if x is positive

$\large [x+1+\frac{1}{x}] \leq -1$ if x is negative

We need minimum value , thats why we will consider the second case. And modulus is applied over the whole expression, so we will take every term (in product) equal to -1.

So min. value $\large =\boxed{|-1|=1}$

$|\frac{(x^2+x+1)(y^2+y+1)(z^2+z+1)(t^2+t+1)}{xyzt}|$

Since all four variables are symmetric, we can write this as $|\frac{x^2+x+1}{x}|^4$

$|\frac{x^2+x+1}{x}|^4= |x+1+\frac{1}{x}|^4$

$|x+1+\frac{1}{x}|^4\ge1^4$ through use of the AM-GM inequality

Thus, our answer is one.

The expression is the product of 4 symmetric expressions. We see that each such expression can be written as (1+x+1/x) which defines a curve with certain minima. Since this minima is same for all 4 expressions, the value of x, y, z & t is same for the minimum value. Thus the expression can be written as (1 + x + 1/x)^4. By calculating the minima of the expression, we find 1 & -1 as the optimum values. It is obvious from the expression that a negative value of x yields lesser value of expression. So putting x = -1, the value of expression becomes (-1)^4 = 1.

simplifying the given data, the result is the absolute value of -1 = 1

minimum of (x+1/x) is at x=-1; hence x+1/x+1=-1; so -1* -1 - 1 -1=1

Plugging in $x=y=z=t=-1$ we see that $1$ is possible.

$0$ is not possible, because $q^2+q+1=0$ does not have a real root.