Quadratic-quadratic minimum

Since $r$ is a non-zero real number and $\frac{1}{r}$ has the same sign as $r,$ $|r+\frac{1}{r}|=|r|+|\frac{1}{r}|=\frac{|r|^2+1}{|r|}\geq \frac{|r|^2-2|r|+1+2|r|}{|r|}\geq \frac{(|r|-1)^2+2|r|}{|r|}\geq \frac{2|r|}{2}\geq 2.$

From the previous inequalities we also obtain that the equality is reached when $r=1$ or $r=-1.$ Now, making $t=r+\frac{1}{r}$ the problem can be rephrased in the following way: we need to find the absolute minimum of the form $(1+t^2)x^2- 6t x+t^2+4$ where $t$ and $x$ are any real numbers and $|t|\geq 2.$

By factoring $(1+t^2)$ out and completing the square we get $(1+t^2)x^2- 6t x+t^2+4=(1+t^2)(x-\frac{3t}{1+t^2})^2+t^2+4-\frac{9t^2}{1+t^2}.\:\:\:\:\:\:\:\:(*)$

Since the term $(1+t^2)(x-\frac{3t}{1+t^2})^2$ is always greater than or equal to zero, we can find the minimum value for the given expression by finding a value of $t$ , such that $|t|\geq 2,$ that minimizes the function $f(t)=t^2+4-\frac{9t^2}{1+t^2},$ and then making $x=\frac{3t}{1+t^2}$ for that value of $t.$ We can verify that $f '(t)= \frac{2 t (t^2-2) (t^2+4)}{(t^2+1)^2},$ so the only roots of the derivative of $f(t)$ are $-\sqrt{2}, 0,$ and $\sqrt{2}.$ and it is never undefined at any number. Now, it is easy to determine by using test points that $f '(t)$ is negative on $(-\infty, -2),$ and positive on $(2 ,\infty ).$ Therefore the minimum value of $f(t)$ for all values of $t$ such that $|t|\geq 2$ will be $f(-2)=f(2)=\frac{4}{5}=0.8.$ Thus we can conclude that the expression $(*)$ gets its minimum value a $t=2$ or $-2$ and taking $x=\frac{3(2)}{1+2^2}=\frac{6}{5}$ or $x=\frac{3(-2)}{1+(-2)^2}=-\frac{6}{5},$ respectively, and the minimum values of $(*)$ will be 0.8.