What is the minimum value of $$f(x) = \sqrt{\frac{2(x - 1)}{x}} + \frac{x + 1}{x},$$ if $x \in \mathbb{R}$ and $x > 1$?
Note that $f$ has a global minimum value of $$f(1) = 2$$ if we allow $x \geq 1$. (The WolframAlpha verification is here.)
A 2-D plot of the function for $x \in (-\infty,\infty)$ is here.
note that $$\frac{d}{dx}f(x)=\frac{d}{dx}\sqrt{\frac{2(x - 1)}{x}} + \frac{x + 1}{x}=\frac{\left(\sqrt{2-\frac2x}-2\right)x+2}{2(x-1)x^2}$$
when $1\lt x\lt2$ , you can see that $f'(x)$ has a positive value and when $x=2$, $f'(x)=0$. when $x\gt2$ $f'(x)$ has a negative value. In fact, it has a max. value at $x=2$ but no min. value given that $x\gt1$. It is not actually strange, not every function has min. value.