# Is there an elegant method to find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$ for positive $x$?

Find the minimum value of $$\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$$ for $$x\gt0$$.

In some textbook, the problem is usually tackled by calculus. So I started to investigate the problem using triangle inequality only. The answer is so interesting and simple that the minimum value of $$S(x)$$ is $$\sqrt{10}$$ when $$x=\frac{1+\sqrt{13}}{6}$$.

My Question: How many elegant methods are there to find the minimum point?

We are going to find the minimum value of} $$S(x)=\dfrac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$$ geometrically using Triangle Inequality.
\displaystyle \begin{aligned} S(x)&=\sqrt{x^{2}+1+\frac{2}{x}+\frac{1}{x^{2}}} +\sqrt{x^{2}-2 x+5-\frac{4}{x}+\frac{1}{x^{2}}} \\&=\sqrt{x^{2}+\left(\frac{1}{x}+1\right)^{2}} +\sqrt{(x-1)^{2}+\left(\frac{1}{x}-2\right)^{2}}\end{aligned} $$\textrm{which is the sum of distances of any point }P\text{ on a rectangular hyperbola from }(0,-1)$$ $$\textrm{and }(1,2) \textrm{ as shown below:}$$
Since $$S(x) =A P+P B \geqslant A B =\sqrt{(1-0)^{2}+(2-(-1))^{2}}=\sqrt{10},$$ therefore the minimum value of S(x) is $$\sqrt{10},\\$$ when and only when A, B and P are collinear$$\quad \textrm{ i.e. } \dfrac{2-\frac{1}{x}}{1-x}=\dfrac{2-(-1)}{1-0 } \Leftrightarrow \dfrac{2 x-1}{1-x}=3 x \Leftrightarrow 3 x^{2}-x-1=0 \Leftrightarrow x=\dfrac{1+\sqrt{13}}{6}.$$
We can now conclude that the minimum value of S(x) is $$\sqrt{10}$$ when $$x=\dfrac{1+\sqrt{13}}{6}.$$