Find the $x$ value for when the expression is takes the minimum value

Question

Let $E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}.$ Find the $x$ value for when this expression take it's minimum value.

Using the derivative it's pretty hard... and I think not indicated for this problem... I thought about using the inequality of means. First, rewrite:

$E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}$ as $E(x)=\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}$..

But I don't know how to continue...

Answer 1

Note that $$ \sqrt{\left(x-1\right)^2+4}=\sqrt{\left(x-1\right)^2+\left(0-\left(-2\right)\right)^2} $$ is exactly the distance between point $A:\left(x,0\right)$ and point $B:\left(1,-2\right)$, and that $$ \sqrt{\left(x-4\right)^2+9}=\sqrt{\left(x-4\right)^2+\left(0-3\right)^2} $$ is exactly the distance between point $A:\left(x,0\right)$ and point $C:\left(4,3\right)$. Therefore, you are to find some point $A$ on the $x$-axis such that $$ d(A,B)+d(A,C) $$ is minimized (i.e., the total distance of the red segments in the figure below is minimized), where $d(A,B)$ denotes the distance between points $A$ and $B$.

Gratefully, it is obvious that, as $A$ falls on the dashed blue segment, the total distance witnesses its minimum, since this dashed blue segment yields the minimum distance between points $B$ and $C$. Due to the triangular inequality, any other position of $A$ would make $$ d(A,B)+d(A,C)>d(B,C). $$ Therefore, the very $A$ that minimizes the total distance should be the point on the dashed blue segment. Then combined with the fact that $A$ is a point on the $x$-axis, this $A$ would have no choice but to be $\left(11/5,0\right)$.

Answer 2

Using Minkowski's inequality: $$\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}=\\ \sqrt{(x-1)^2+2^2}+\sqrt{(4-x)^2+3^2}\ge \\ \sqrt{(x-1+4-x)^2+(2+3)^2}=\sqrt{3^2+5^2}=\sqrt{34}.$$ The equality occurs when $(x-1,2)||(4-x,3)$, that is: $$\frac{3}{2}(x-1)=4-x \Rightarrow x=\frac{11}{5}.$$