$\min (y)$

Make the substitution $x=-z$ to get

$y=\sqrt{(-z\sqrt{5}+1)^2}-z\sqrt{5}$

$y=\sqrt{(z\sqrt{5}-1)^2}-z\sqrt{5}$

$y=(z\sqrt{5}-1)-z\sqrt{5}$ for $z\sqrt{5} \ge 1$

$y=-1$ for $x \le -\dfrac{1}{\sqrt{5}}$

Given that

$\begin{aligned} y & = \sqrt{5x^2+2x\sqrt 5+1} + x\sqrt 5 \\ & = \sqrt{({\color{#3D99F6}x\sqrt 5+1})^2} + x\sqrt 5 \end{aligned}$

For $y$ to be real, $\color{#3D99F6}x\sqrt 5 + 1 \ge 0$ , $\implies x \ge - \frac 1{\sqrt 5}$ and $y$ is minimum when $x = - \frac 1{\sqrt 5}$ that is $\min(y) = \sqrt 0 + \left(-\frac 1{\sqrt 5}\right) \sqrt 5 = \boxed{-1}$ .

I did a long-winded calculus approach so bare with me!

Since we are trying to find the minimum value, we know that we can find the derivative of the function and find where it is equal to zero

$y = \sqrt{5x^{2}+2x\sqrt{5}+1}+x\sqrt{5}$

$\frac{dy}{dx}=\frac{10x+2\sqrt{5}}{2\sqrt{5x^{2}+2x\sqrt{5}+1}}+\sqrt{5} = \frac{5x+\sqrt{5}}{\sqrt{5x^{2}+2x\sqrt{5}+1}}+\sqrt{5}$

Now if we set it equal to zero, and using some algebra...

$\frac{5x+\sqrt{5}}{\sqrt{5x^{2}+2x\sqrt{5}+1}}+\sqrt{5} = 0$

$\frac{5x+\sqrt{5}}{\sqrt{5x^{2}+2x\sqrt{5}+1}} = -\sqrt{5}$

$5x+\sqrt{5} = -\sqrt{5}\sqrt{5x^{2}+2x\sqrt{5}+1}$

$5x+\sqrt{5} = -\sqrt{5(5x^{2}+2x\sqrt{5}+1)}$

$5x+\sqrt{5} = -\sqrt{25x^{2}+10x\sqrt{5}+5}$

$5x+\sqrt{5} = -\sqrt{( 5x+\sqrt{5})^{2}}$

$5x+\sqrt{5} = -5x-\sqrt{5}$

Solving for x, we get... $x = \frac{-\sqrt{5}}{5}$

Finally, we just have to plug it into the original equation, and we get

$y = \sqrt{5(\frac{-\sqrt{5}}{5})^{2}+2(\frac{-\sqrt{5}}{5})\sqrt{5}+1}+(\frac{-\sqrt{5}}{5})\sqrt{5}$

$y = 0 + (\frac{-\sqrt{5}}{5})\sqrt{5}$ )

$y = \boxed{-1}$

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