<s>[INST] Determining the minimum value of the function $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ I am curious whether there is an algebraic verification for $y = x + 2\sqrt{x^2 - \sqrt{2}x + 1}$ having its minimum value of  $\sqrt{2 + \sqrt{3}}$ at $\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}$. I have been told the graph of it is that of a hyperbola.
 [/INST] Here is a "non-calculus" way that plays the whole show back to AMGM. It is a bit cumbersome but works.
I prefer giving all stepwise substitutions to show how to bring the whole expression back to hyperbolic functions where AMGM suddenly gives all. The basic idea behind it is that $\cosh t = \sqrt{\sinh^2 t + 1}$:
$$\color{blue}{y=} x + 2\sqrt{x^2 - \sqrt{2}x + 1}$$
$$x^2 - \sqrt{2}x + 1 = (x - \frac{\sqrt{2}}{2})^2 + 1 - \frac{1}{2} = (x - \frac{\sqrt{2}}{2})^2 +  \frac{1}{2}$$
$$\color{green}{u =x - \frac{\sqrt{2}}{2}}  \Rightarrow \color{blue}{y=} u + \frac{\sqrt{2}}{2} +2\sqrt{u^2+\frac{1}{2}} = \color{blue}{\frac{\sqrt{2}}{2} + u + \sqrt{2}\sqrt{\left(\sqrt{2} u\right)^2 + 1}}$$
$$\color{green}{v = \sqrt{2} u} \Rightarrow \color{blue}{y =} \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} v + \sqrt{2}\sqrt{v^2 + 1} = \color{blue}{\frac{\sqrt{2}}{2} + \sqrt{2} \left(\boxed{ \frac{v}{2} + \sqrt{v^2 + 1}} \right)}$$
$$\color{green}{v = \sinh t} \Rightarrow \boxed{ \frac{\sinh t}{2} + \cosh t} = \frac{e^t - e^{-t}}{4} + \frac{e^t + e^{-t}}{2} = \frac{3}{4}e^t + \frac{1}{4} e^{-t}  \boxed{\stackrel{AMGM}{\geq}} \sqrt{\frac{3}{4}} = \boxed{\frac{\sqrt{3}}{2}}$$
Setting $\color{green}{w= e^t}$, equality holds for
$$3w = \frac{1}{w} \stackrel{w>0}{\Leftrightarrow} w =\frac{1}{\sqrt{3}} \Rightarrow 
\color{blue}{y \geq} \frac{\sqrt{2}}{2} + \sqrt{2}\frac{\sqrt{3}}{2} = \color{blue}{\frac{\sqrt{2}}{2}\left( 1 + \sqrt{3}\right)}$$
Note that
$$\left( \frac{\sqrt{2}}{2}\left( 1 + \sqrt{3}\right) \right)^2 =\frac{1}{2}\left( 1 + 2\sqrt{3} + 3\right) =  2 + \sqrt{3} $$
Backwards substitution yield $x$:
$$\color{green}{t =} \ln \frac{1}{\sqrt{3}} = \color{green}{-\ln \sqrt{3}} \Rightarrow \color{green}{v =} \sinh t = \frac{\frac{1}{\sqrt{3}} - \sqrt{3}}{2} = \color{green}{-\frac{\sqrt{3}}{3}}$$ $$ \Rightarrow \color{green}{x =} \frac{\sqrt{2}}{2}v+\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \cdot \left(- \frac{1}{\sqrt{3}} \right) + \frac{1}{\sqrt{2}} = \color{green}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}}$$
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