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.
 A: Without derivatives,
Considering the functions
$$
f(x,y)=y-x-2\sqrt{x^2-\sqrt 2 x-1} = 0\\
y = \lambda
$$
Their intersection is at the solution for
$$
f(x,\lambda)=\lambda-x-2\sqrt{x^2-\sqrt 2 x-1} = 0
$$
or squaring
$$
(x-\lambda)^2-2(x^2-\sqrt 2 x-1)=0
$$
Solving for $x$ we have
$$
x = \frac{1}{3} \left(2 \sqrt{2}\pm 2 \sqrt{\lambda ^2-\sqrt{2} \lambda -1}-\lambda\right)
$$
but at tangency between $f(x,y)=0, y =  \lambda$ we have
$$
\lambda ^2-\sqrt{2} \lambda -1 = 0\Rightarrow \lambda = \frac{1}{2} \left(\sqrt{2}+\sqrt{6}\right)
$$
as the feasible minimum.

A: One of the approaches may be as follows:
Suppose there exists some $a$ which is the minimum. Then:
$$
\begin{align}
x + 2\sqrt{x^2 - \sqrt2 x+1} &= a \\
2\sqrt{x^2 - \sqrt2 x+1} &= a-x
\end{align}
$$
Square both sides:
$$
(a-x)^2 = 4x^2 - 4\sqrt2x + 4
$$
Applying some transformations you can get:
$$
3x^2 + x(2a-4\sqrt2) + 4 -a^2 =0
$$
Now you want the discriminant to be equal to zero which means only one root will exist, so:
$$
D = 16a^2 - 16\sqrt2a-16 = a^2 - \sqrt2a - 1 =0
$$
The equation in terms of $a$ has two solutions:
$$
a_1 = {1-\sqrt3 \over \sqrt2} \\
a_2 = {1+\sqrt{3} \over \sqrt2}
$$
If you plug them into the initial equation only one of them is going to produce a valid statement. Therefore that will be you minimum, which appears to be ${1+\sqrt3 \over \sqrt2}$
A: 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 $x$:
$$\color{green}{t =} \ln \frac{1}{\sqrt{3}} = \color{green}{-\ln \sqrt{3}} \Rightarrow \color{green}{v =} \sinh t = \frac{\frac{1}{\sqrt{3}} - \sqrt{3}}{2} = \color{green}{-\frac{\sqrt{3}}{3}}$$ $$ \Rightarrow \color{green}{x =} \frac{\sqrt{2}}{2}v+\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \cdot \left(- \frac{1}{\sqrt{3}} \right) + \frac{1}{\sqrt{2}} = \color{green}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}}$$
A: 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
$$\begin{align}x&=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}&y&=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\text{.}
\end{align}$$
