For real numbers $x$ and $y$ , find the minimum value of $\sqrt{x^2+y^2}+\sqrt{(x-3)^2+(y-4)^2}$

Question

Thailand International Math Olympiad 2018 secondary senior problem 18

For real numbers $x$ and $y$ , find the minimum value of $$ \sqrt{x^2 + y^2} + \sqrt{(x-3)^2 + (y-4)^2} $$

This question was under geometry. I found this really challenging. I couldn't see any geometry other than circle equations in this problem.

Answer 1

Note that the formulas seem similar to the distance formula in $\mathbb R^2$. In fact, the formula is the sum of the distances of point $(x,y)$ to $(0,0)$ and $(3,4)$. We know that any point on the segment in between $(0,0)$ and $(3,4)$ will minimize the function by the triangle inequality. In particular, $(0,0)$ will.

So, we simply evaluate $$\sqrt{3^2+4^2}=5$$To get our answer of $5$.

Answer 2

Since it is a senior secondary problem one may surely use the Minkowski inequality:

We have

$$\sqrt{x^2+y^2}+\sqrt{(x-3)^2+(y-4)^2}=\sqrt{x^2+y^2}+\sqrt{(3-x)^2+(4-y)^2}$$ $$\stackrel{Minkowski}{\geq} \sqrt{(3-x+x)^2 + (4-y+y)^2}=\sqrt{25}=5.$$

Equality is achieved for $\binom xy = \lambda \binom{3-x}{4-y}$, for $\lambda \geq 0$ or $\binom{3-x}{4-y}=\binom 00$. So, the lower bound is attained, for example, for $x=3, y=4$.