Let $a>b>0$.
Suppose we want to minimize $$ f(x)=(x-a)^2+(1/x-b)^2, $$ over $x>0$.
Equating $f'(x)=0$ leads to the quartic equation $$ x^4-ax^3+bx-1=0. $$ Is there any way to determine which$$ g(x)=x^4-ax^3+bx-1=0. \tag{1} $$
Question:
Is the smallest positive real root of the equation is$(1)$ always the right oneminimizer of $f$? i.e
Since $g(0)<0$ and $\lim_{x \to -\infty} g(x)=\lim_{x \to \infty} g(x)=\infty$, there always exist two real solutions, one positive and one negative. if we order the r
Thus, there can be either one positive root (e.g. smallest/largestwhen $a=b=1$).
I tried various things, eor three positive roots (e.g. considering the expression $f(x)-f(\tilde x)$ when $x, \tilde x$ are roots, bot got essentially nowhere$a=3,b=4$).
In particularsome numerical examples I tried, at any given rootthe smallest $x$(positive) root was indeed the minimizer, $$ f(x)=(x-a)^2(1+x^4), $$ but and I wonder whether this does not seem to help muchis always the case.