<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] 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 ${1+\sqrt3 \over \sqrt2}$
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