Finding minimum value of $\sqrt{x^2-2x+1} + \sqrt{x^2-6x+13}$

Question

How can I minimize $f$, where $$f(x)=\sqrt{x^2-2x+1} + \sqrt{x^2-6x+13}$$

I know how to minimize a quadratic equation. But I don't know how to minimize a case like this where two quadratics are involved and that too inside a square root function.

Answer 1

By Minkowski (triangle inequality) we obtain: $$\sqrt{x^2-2x+1}+\sqrt{x^2-6x+13}=\sqrt{(x-1)^2+0^2}+\sqrt{(3-x)^2+2^2}\geq$$ $$\geq\sqrt{(x-1+3-x)^2+(0+2)^2}=2\sqrt2.$$ The equality occurs for $x=1$, which says that $2\sqrt2$ is a minimal value.

Similarly we obtain: $$\sqrt{x^2-2x+2}+\sqrt{x^2-6x+13}=\sqrt{(x-1)^2+1^2}+\sqrt{(3-x)^2+2^2}\geq$$ $$\geq\sqrt{(x-1+3-x)^2+(1+2)^2}=\sqrt{13}.$$ The equality occurs for $(x-1,1)||(3-x,2)$, id est, for $x=\frac{5}{3}$.

Answer 2

I assume you know how to complete the square, to find where the quadratics are minimized. Each of the two summands in your $f$ is piecewise concave. (Graph them to see what I mean.) The first term, $|x-1|$, is concave on $(-\infty,1]$ and concave on $[1,+\infty)$. The second is concave on $(-\infty,3]$ and concave on $[3,+\infty)$. So your $f$ is concave on each of the intervals $(\infty,1]$, $[1,3]$, and $[3,+\infty).$ Now use the fact that a concave function can attain its minimum on an interval only at an endpoint. That gives you 4 endpoints to check ($\pm\infty$, $1$, and $3$); the one that gives the smallest $f$ is what you are looking for.

Answer 3

To minimize $f(x)=\sqrt{(x^2-2x+2)} + \sqrt{(x^2-6x+13)}$, you can do this directly by taking the derivative of $f$ with respect to $x$: $$ \color{green}{f'(x)=\frac{2 x-2}{2 \sqrt{x^2-2 x+2}} + \frac{2 x-6}{2 \sqrt{x^2-6 x+13}}} $$ and set it equal to zero: $\color{blue}{f'(x)=0}$. Now, solve for $x$: $$ \frac{2 x-2}{2 \sqrt{x^2-2 x+2}} =- \frac{2 x-6}{2 \sqrt{x^2-6 x+13}} $$ implies $$ (x-1)\sqrt{x^2-6 x+13} =- (x-3)\sqrt{x^2-2 x+2} \hspace{8mm} (\spadesuit) $$ by clearing the denominators. Square both sides to get: $$ (x-1)^2 (x^2-6 x+13) =(x-3)^2 (x^2-2 x+2) \hspace{8mm} (\diamondsuit) $$ or $$ 3 x^2-2 x-5=(x+1)(3x-5)=0. $$ So $x=-1$ or $x=5/3$. Since $$ f(-1) = 1 + 2 \sqrt{2} \cong 3.82843 \hspace{4mm}\mbox{ and }\hspace{4mm} f(5/3) = \sqrt{13}\cong 3.60555, $$ we conclude that $f$ attains its minimum when $\color{magenta}{\boxed{x=5/3}}$.

$\textbf{Remark}.$ Note that $x=-1$ is $\textit{not}$ a significant point of $f$. It appeared as it if were a critical point when we went from $(\spadesuit)$ to $(\diamondsuit)$ by squaring both sides. As a further analysis, here is a plot of
$g(x)=(x-1)\sqrt{x^2-6 x+13} + (x-3)\sqrt{x^2-2 x+2}$: and here is a plot of $h(x) = (x-1)^2 (x^2-6 x+13) -(x-3)^2 (x^2-2 x+2)$: .