<s>[INST] minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$ 
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of  $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$

what i try 
If $x\neq 0,$ Then  divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\displaystyle x+\frac{1}{x}=t,$ then $\displaystyle x^3+\frac{1}{x^3}=t^3-3t$
$$f(t)=\frac{(t-1)^3}{t^3-3t}$$
How do i solve it without derivatives
Help me please 
 [/INST] A solution without derivatives
Clearly, $t \ge 2$ or $t \le -2$. 
Denote the minimum of $f(t)=\frac{(t-1)^3}{t^3-3t-1}$ on $|t|\ge 2$ by $m > 0$.
Since $f(3) = \frac{8}{17}$, we have $m \le \frac{8}{17}$.
Note that $f(t) - 1 = \frac{3t(2-t)}{t^3-3t-1} \ge 0$ for $t \le -2$.
Thus, the minimum occurs on $t \ge 2$.
$f(t) = m$ leads to a cubic equation $g(t) = 0$ where
\begin{align}
g(t) &= (t^3-3t-1)(f(t)-m)\\
&= (1-m)t^3 - 3t^2 + (3+3m)t + m - 1.
\end{align}
Clearly, $g(t) = 0$ has a real root on $t \ge 2$.
Since $g(0) = m - 1 < 0$ and $g(2) = 1 - m > 0$, we know that $g(t) = 0$ has a real root on $(0, 2)$.
Thus, we can factor $g(t) = (1-m)(t-t_1)(t-t_2)(t-t_3)$ for some $t_1 \in (0, 2)$ and $t_2 \ge 2$.
Since $g(t) \ge 0$ for $t \ge 2$, it is easy to prove that $t_2 = t_3$.
Thus, we have $g(t) = (1-m)(t-t_1)(t-t_2)^2$. 
Since $g(t) = 0$ has a multiple root, its discriminant is zero, 
i.e. $\Delta = 81(m^2+6m-3)m^2 = 0$ which results in $m = 2\sqrt{3} - 3$.
Finally, let us prove that $m = 2\sqrt{3} - 3$ is indeed the minimum of $f(t)$ on $|t|\ge 2$.
It is easy to check that
$$f(t) - (2\sqrt{3} - 3) = \frac{1}{t^3 - 3t - 1}(4-2\sqrt{3})\Big(t - 1 + \tfrac{\sqrt{3}}{2}\Big)\Big(t - 1 - \sqrt{3}\Big)^2$$
and hence $f(t) - (2\sqrt{3} - 3) \ge 0$ for $|t| \ge 2$, with equality if and only if $t = 1 + \sqrt{3}$.
We are done.
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