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 root of the equation is the right one? i.e. if we order the r (e.g. smallest/largest).

I tried various things, e.g. considering the expression $f(x)-f(\tilde x)$ when $x, \tilde x$ are roots, bot got essentially nowhere.

In particular, at any given root $x$, $$ f(x)=(x-a)^2(1+x^4), $$ but this does not seem to help much.