# Showing that the Ratio of Two Expressions is Always Greater Than 1

• brainpushups
In summary, the conversation discusses a proof by Huygens about comparing two expressions and finding the maximum value of a ratio. The proof involves using line segments and areas of rectangles. An algebraic approach is also suggested and shown to be valid.
brainpushups
I was trying to follow a proof that Huygens wrote about collisions and was having a hard time deciphering his rather lengthy geometric presentation. Essentially I want to compare two expressions

$$\frac{2x}{x+z}$$
and
$$\frac{4xy}{(x+y)+(y+z)}$$

Where, for the sake of this thread let's say that ##x<y<z## But really all that Huygens requires is that y is an intermediate value; x does not need to be the smallest value.

What is to be shown is that the second expression is always larger than the first. Furthermore, it is at its largest when the value of y is the geometric mean of x and z. I chose to take the ratio of the expressions to get

$$\frac{(x+y)(y+z)}{2y(x+z)}$$

I guess I don't have much experience with this sort of thing; I'm not even sure how to go about the first part which perhaps seems rather trivial (or maybe not). Just to be clear - I'd like to show that the fraction above is always greater than 1 for ##x<y<z## and that this ratio is greatest when ##y=\sqrt{xz}## Can anybody offer either some insight or a suggestion of what to further study that would help me gain some experience for how to go about this? Huygens does it by using line segments to compare quantities and ultimately comparing areas formed by rectangles of the segments but, as I said, it was challenging to follow in detail.

I will assume that ##0<x<y<z##.

Let ##f(y) = \frac{(x+y)(y+z)}{2y (x+z)}##. Then it is easily seen that
$$f^\prime(y) = \frac{y^2 - xz}{2y^2(x+z)}$$
We see that ##f^{\prime}(y) = 0## if and only if ##y = \sqrt{xz}##. This could be a local minimum or a local maximum. To see which, we find the second derivative:
$$f^{\prime\prime}(y) = \frac{xz}{y^3(x+z)}$$. This is always positive, so we have a local minimum.

Furthermore, since ##f^\prime(y)## only has one (positive) zero, it will be a global minimum on ##\mathbb{R}^+##.

On ##[x,z]##, the maxima will be attained in either ##x## or ##z##. We see that ##f(x) = 1## and ##f(z) = 1##. Thus we see that ##f(y)<1## for each ##y\in (x,z)##.

brainpushups
Thanks for the reply. I can't believe it didn't occur to me to do that! I think after trying to decipher Huygens for half an hour I was resigned to use only the tools available to him! Just out of curiosity: can you think of a way to show this from a strictly algebraic approach?

To show that
$$\frac{(x+y)(y+z)}{2y(x+z)}< 1,$$
it suffices to show that
$$(x+y)(y+z) < 2y(x+z)$$
For this, it suffices to show that
$$xy + xz + yz + y^2 < 2xy +2yz$$
which is equivalent with
$$xz + y^2 < xy + yz$$
or
$$y^2 -(x+z)y + xz <0$$
This is equivalent to
$$(y-x)(y-z)<0$$
This is indeed true if ##y>x## and ##z>y##.

Now we wish to show that for all ##y>0##, we have
$$\frac{(x+\sqrt{xz})(\sqrt{xz} + z)}{2\sqrt{xz}(x+z)} \leq \frac{(x+y)(y+z)}{2y(x+z)}$$
which is equivalent to
$$(yx + yz)(x\sqrt{xz} + z\sqrt{xz} +2xz)\leq (x\sqrt{xz} + z\sqrt{xz})(xy + xz +yz + y^2)$$
which is equivalent to
$$yx^2\sqrt{xz} + 2xyz\sqrt{xz} +yz^2\sqrt{xz}+ 2x^2yz + 2xyz^2\leq x^2y\sqrt{xz} + x^2z\sqrt{xz} + 2xyz\sqrt{xz} + xy^2\sqrt{xz} + xz^2\sqrt{xz} + yz^2\sqrt{xz} + y^2z\sqrt{xz}$$
This reduces to
$$2x^2yz + 2xyz^2\leq x^2z\sqrt{xz} + xy^2\sqrt{xz} + xz^2\sqrt{xz}+y^2z\sqrt{xz}$$
or
$$(x+z)\sqrt{xz}y^2 -2xz(x + z)y + xz(x+z)\sqrt{xz}\geq 0$$
or
$$\sqrt{xz}y^2 -2xzy + xz\sqrt{xz}\geq 0$$
This is just
$$\sqrt{xz}(y - \sqrt{xz})^2 \geq 0$$
which is obviously true.

Last edited:
brainpushups

## 1. What is the purpose of comparing two expressions?

The purpose of comparing two expressions is to determine if they are equivalent or if one is greater than, less than, or equal to the other. This is useful in many areas of science, such as in data analysis or when solving mathematical equations.

## 2. How do I compare two expressions?

To compare two expressions, you must first simplify each expression as much as possible. Then, you can use mathematical operations like addition, subtraction, multiplication, and division to see if the two expressions are equal or if one is larger than the other. You can also use logical operators like "equal to," "less than," or "greater than" to compare the expressions.

## 3. Can I compare expressions with variables?

Yes, you can compare expressions with variables. In fact, comparing expressions with variables is often necessary in scientific experiments and equations. Just be sure to use the same rules as when comparing numerical expressions, such as simplifying and using mathematical and logical operations.

## 4. What is the difference between numerical and algebraic expressions?

Numerical expressions contain only numbers and mathematical operations, while algebraic expressions contain variables as well as numbers and mathematical operations. When comparing numerical expressions, you are comparing specific numerical values. When comparing algebraic expressions, you are determining if the expressions are equivalent for any values of the variables.

## 5. Why is it important to compare two expressions in science?

Comparing two expressions is important in science because it allows us to understand and analyze data, equations, and relationships between variables. By comparing expressions, we can make predictions, draw conclusions, and further our understanding of the world around us.

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