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.

Scale and shift
$$\begin{align}u&=\sqrt{2}x-1&v&=\tfrac{1}{\sqrt{2}}y-\tfrac{1}{2}\text{.}
\end{align}$$
Then it is just as well to minimize
$$v=\tfrac{u}{2}+\sqrt{u^2+1}\text{.}$$
with respect to $u$.
Use the stereographic projection
$$u=\frac{2t}{t^2-1}\text{.}$$
Then it is just as well to minimize
$$v=1+\frac{t+2}{t^2-1}=1+\frac{3}{2(t-1)}-\frac{1}{2(t+1)}$$
over $t^2>1\text{.}$
Use the Cayley transform
$$s=\frac{t-1}{t+1}\text{.}$$
Then it is just as well to minimize
$$v=\frac{3}{4s}+\frac{s}{4}$$
over $s>0$. Rescale
$$\begin{align}w&=2\frac{v}{\sqrt{3}}&r&=\frac{s}{\sqrt{3}}\text{.}\end{align}$$
Then it is just as well to minimize
$$w=\frac{r+\tfrac{1}{r}}{2}$$
over $r>0$.
But by the arithmetic-geometric mean inequality,
$$w\geq 1$$
with equality iff $r=1$. Retracing the steps gives
$$\begin{align}x&=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}&y&=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\text{.}
\end{align}$$
