How to find min of $\sqrt{x^2+4x+8}+\sqrt{x^2-6x+10}$ without differentiation

Question

I want to find minimum of $$\sqrt{x^2+4x+8}+\sqrt{x^2-6x+10}$$ without differentiation. Although I did it with differentiation and $x=4/3$ then find $f(\frac 43)$ but I am unable to find the min without differentiation.
Implicit : I did this $$\frac{x^2+4x+8-(x^2-6x-10)}{\sqrt{x^2+4x+8}-\sqrt{x^2-6x+10}}$$but get nothing, because the $\sqrt{x^2+4x+8}+\sqrt{x^2-6x+10}>5$
now, If someone help me find it, or gives me a clue I appreciate this.

Answer 1

As pointed out in comments, this is equivalent to minimizing the distance from $P=(3, 1)$ to the x-axis to $Q=(-2, 2)$.

The distance is minimized when the angle of incidence equals the angle of reflection*. So if the points were of the form $(a, b)$ and $(-ak, bk)$, the reflection point that minimizes would be at the origin. So when $P_x / P_y = -Q_x / Q_y$, the origin minimizes the distance.

So shift $P$ and $Q$ left so that they make equal angle with the origin and x-axis.

The shift amount, $t$, then must satisfy $$\frac{Px - t}{Py} = -\frac{Qx - t}{Qy}$$

So you get $$t=\frac{P_xQ_y+P_yQ_x}{Py+Qy} = \frac{3 \times 2 + 1 \times -2}{1 + 2} = \frac{4}{3}$$

*Both points have to be on the same side of the x-axis for this to work. If they are on opposite sides of the x-axis, this is equivalent to the concept "the shortest distance between two points is a straight line".