# Comparing Two Expressions

1. Aug 14, 2015

### 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.

2. Aug 14, 2015

### micromass

Staff Emeritus
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 $y\in (x,z)$.

3. Aug 15, 2015

### 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?

4. Aug 15, 2015

### micromass

Staff Emeritus
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: Aug 15, 2015