Minkowski's inequality and a sum of two square roots

1. Let $u=(p_1,q_1)$ and $v=(p_2,q_2)$. By Cauchy–Schwarz, $u\cdot v\le|u||v|$, so
$$|u+v|^2=|u|^2+2u\cdot v+|v|^2\le|u|^2+2|u||v|+|v|^2=(|u|+|v|)^2.$$
Taking square roots gives the inequality. Equality iff $u,v$ are parallel with the same direction.

2. Write the expression as $\sqrt{x^2+2^2}+\sqrt{(6-x)^2+3^2}$ and apply part 1 with $(p_1,q_1)=(x,2)$ and $(p_2,q_2)=(6-x,3)$:
$$\ge\sqrt{6^2+5^2}=\sqrt{61}.$$

Equality. The vectors $(x,2)$ and $(6-x,3)$ must be parallel with the same direction: $3x=2(6-x)$, i.e. $x=\frac{12}5>0$. So the minimum is $\sqrt{61}$, reached at $x=\frac{12}5$.

Remark. Geometrically, the expression is the length of the broken line from $(0,2)$ to $(x,0)$ to $(6,-3)$; it is shortest when the line is straight, of length $\sqrt{6^2+5^2}$.

Lesson. A sum of square roots of sums of two squares is a sum of lengths: think "triangle inequality", and choose the vectors so that the unknown cancels in the sum.