# Min Max problem: the shortest distance for a light ray

• Karol
In summary, the attempt at a solution is to use GM≤AM to prove that if the sum of squares of the distances (setup 2) in an arbitrary point is bigger than the sum of the squares of the shortest distances (setup 1, the mid point between A and B), then the distance from an arbitrary point is also bigger.

## Homework Equations

Minimum/Maximum occurs when the first derivative=0
GM≤AM: ##~\sqrt{xy}\leq\frac{x+y}{2}##

## The Attempt at a Solution

[/B]
If the sum of squares of the distances (setup 2) in an arbitrary point is bigger than the sum of the squares of the shortest distances (setup 1, the mid point between A and B) then the distance from an arbitrary point is also bigger. i want to prove without calculus that:
$$2s^2+x^2+y^2\geq 2s^2+2\left[ \frac{x+y}{2} \right]~\rightarrow~\frac{3}{2}(x^2+y^2)\geq xy$$
I want to use GM≤AM so i take the square root of what i need to prove, and i need to prove:
$$\sqrt{1.5}\sqrt{x^2+y^2}\geq\sqrt{xy}$$
I know ##~\sqrt{xy}\leq\frac{x+y}{2}~## so i have to prove:
$$\sqrt{xy}\leq\frac{x+y}{2}\leq\sqrt{1.5}\sqrt{x^2+y^2}$$
And it remains that i have to prove:
$$\frac{x+y}{2}\leq\sqrt{1.5}\sqrt{x^2+y^2}$$
I know ##~\sqrt{(x+y)^2}=\sqrt{x^2+y^2+2xy}\leq\sqrt{x^2+y^2}~## but it doesn't help since i need to insert an inequality
on the left of ##~\sqrt{1.5}\sqrt{x^2+y^2}~## and not on the right side.

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You can use AM-QM

AM-QM, in my case, is exactly GM≤AM: ##~~\sqrt{xy}\leq\frac{x+y}{2}##
It doesn't help in proving ##~\frac{x+y}{2}\leq\sqrt{1.5}\sqrt{x^2+y^2}##

Karol if you want to prove

##\frac{3}{2}(x^2+y^2)\geq xy##

first recognize that ##\frac{3}{2}(x^2+y^2) \geq \frac{1}{2}(x^2+y^2)##

now ##xy = \big(x^2 y ^2\big)^{\frac{1}{2}}= \sqrt{\big(x^2 y ^2\big)}##... right? What can you do with that?

Karol said:
I know ##\sqrt{x^2+y^2+2xy}\leq\sqrt{x^2+y^2}~##
You do? Try ##x=y=1##

$$\frac{3}{2}(x^2+y^2)\geq \frac{x^2+y^2}{2}\geq \sqrt{x^2y^2}=xy$$
But i am not sure i can use the same x for the two setups, 1 and 2.

I want the shortest distance between A and B. AC is the shortest and equals the shortest AB

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## 1. What is a "Min Max problem" in relation to light rays?

A Min Max problem is a mathematical optimization problem that aims to minimize the maximum value of a given function. In the context of light rays, this problem involves finding the shortest distance that a light ray can travel from one point to another while encountering various obstacles or reflecting off different surfaces.

## 2. How is the shortest distance for a light ray calculated in a Min Max problem?

The shortest distance for a light ray is calculated by finding the minimum value of the maximum distance that the light ray can travel while encountering different obstacles or reflecting off surfaces. This is achieved through various mathematical techniques, such as gradient descent, linear programming, or dynamic programming.

## 3. What factors affect the shortest distance for a light ray in a Min Max problem?

The factors that affect the shortest distance for a light ray in a Min Max problem include the initial and final positions of the light ray, the properties of the obstacles or surfaces it encounters, and the angle at which it reflects off these surfaces. Other factors may also include the refractive index of the medium through which the light ray is traveling.

## 4. How is the Min Max problem for light rays used in real-life applications?

The Min Max problem for light rays has various real-life applications, such as in computer graphics, where it is used to calculate the shortest distance for light rays to travel in order to create realistic reflections and shadows. It is also used in the design of optical systems, such as lenses and mirrors, to ensure that the maximum amount of light is collected or focused.

## 5. What are the limitations of using the Min Max problem for light rays?

One limitation of using the Min Max problem for light rays is that it assumes a perfect environment with no imperfections or variations in the properties of the surfaces or obstacles. In reality, these imperfections may affect the actual shortest distance that a light ray can travel. Additionally, the computational complexity of solving the Min Max problem may increase significantly as the number of obstacles or surfaces increases, making it difficult to find an exact solution in a timely manner.