Find minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$

Question

Find minimum value of $f(x) $ where $$f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$

On differentiating I got $$f'(x)=\frac{3(x^2-x+1)^2\left(x^6-2x^5-x^4+x^2+2x-1\right)}{(x^6-x^3+1)^2}$$ which doesn't help much. There could be another way

Answer 1

Divide numerator and denominator by $x^3$

Consider then $x+\frac{1}{x}=t$.

Moreover, $x^3+\frac{1}{x^3}=(x+\frac{1}{x})((x+\frac{1}{x})^2-3)$

Answer 2

Well it's a very complicated way but let me propose it because it's general :

The problem is equivalent to :

$$f(x)=\frac{g(x)^3}{g(x^3)}$$

Where $g(x)=x^2-x+1$

We derivate it to get :

$$f'(x) = \frac{(3 g^2(x) (g(x^3)g'(x)-x^2g(x)g'(x^3))}{g^2(x^3)}=0\quad (1)$$

Now derivate :

$$h(x)=\ln\Big(f(x)\Big)$$

In term of the function $g(x)$ we get :

$$h'(x)=-\frac{3 (x^2 g(x) g'(x^3) - g(x^3) g'(x)))}{(g(x) g(x^3))}=0\quad(2)$$

So we see that in $(1)$ and $(2)$ there is an equivalence concerning the minimum.

So we want the minimum value of $h(x)$ now I get (derivating+multiply by the two denominators+factorize):

$$3(x-1)(x+1)(x^4-2x^3-2x+1)=0$$

See WA to conclude

Answer 3

The numerator of your expression for the derivative is in fact $$3 (x-1) (x+1) \left(x^2-x+1\right)^2 \left(x^4-2 x^3-2 x+1\right)$$ $\left(x^2-x+1\right)=0$ does not show real roots and for the quartic $\left(x^4-2 x^3-2 x+1\right)=0$, using the method given here, $\Delta=-1728$ then the equation has two distinct real roots and two complex conjugate non-real roots.

Continuing with the method described in the linked page, then the real roots $$x_1=\sqrt{1+\sqrt{3}-\sqrt{3+2 \sqrt{3}} } \qquad x_2=\sqrt{1+\sqrt{3}+\sqrt{3+2 \sqrt{3}} }$$