<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] 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.

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