jmc

We can write $$\begin{array}{rl}f(x)& =\sqrt{5x^2+2x\sqrt{5}+1}+x\sqrt{5}\\ & =\sqrt{(x\sqrt{5}+1{)}^2}+x\sqrt{5}\\ & =|x\sqrt{5}+1|+x\sqrt{5}.\end{array}$$If $x\le -\frac{1}{\sqrt{5}},$ then $$f(x)=|x\sqrt{5}+1|+x\sqrt{5}=-x\sqrt{5}-1+x\sqrt{5}=-1.$$If $x\ge -\frac{1}{\sqrt{5}},$ then $$\begin{array}{rl}f(x)& =|x\sqrt{5}+1|+x\sqrt{5}\\ & =x\sqrt{5}+1+x\sqrt{5}\\ & =(x\sqrt{5}+1)+(x\sqrt{5}+1)-1\\ & \ge -1.\end{array}$$Thus, the minimum value of $f(x)$ is $\boxed{-1}.$