# Prove of new metric space

1. Mar 18, 2012

### Lily@pie

Suppose that (X,d) is a metric

Show $\tilde{d}$(x,y) = $\frac{d(x,y)}{\sqrt{1+d(x,y)}}$ is also a metric

I've proven the positivity and symmetry of it.

Left to prove something like this

Given a$\leq$b+c
Show $\frac{a}{\sqrt{1+a}}$$\leq$$\frac{b}{\sqrt{1+b}}$+$\frac{c}{\sqrt{1+c}}$

I try to prove this
a$\sqrt{1+b}$$\sqrt{1+c}$=b$\sqrt{1+a}$$\sqrt{1+c}$+c$\sqrt{1+a}$$\sqrt{1+b}$
but i'm just stuck!

cz previously i've proven this $\frac{a}{1+a}$$\leq$$\frac{b}{1+b}$+$\frac{c}{1+c}$ before....

I guess I can't use the same method??

2. Mar 18, 2012

### Norwegian

Can you prove this assuming a is bigger than both b and c? Can you prove it assuming a is not the biggest?

3. Mar 18, 2012

### Lily@pie

What do you mean? I know a≤b+c...

So I assume a is bigger than both a and c?

4. Mar 19, 2012

### Norwegian

I mean that if you can prove an assertion under some condition, and then under the complementary condition, then you are done. And I have also told you what condition works well in your case.

5. Mar 19, 2012

### Jamma

This is natural to assume given the triangle inequality for the original metric.

6. Mar 19, 2012

### Lily@pie

Let:
a=d(x,y)
b=d(x,z)
c=d(z,y)

a$\leq$b+c
This is the triangle inequality for the original metric right?

Show $\frac{a}{\sqrt{1+a}}$$\leq$$\frac{b}{\sqrt{1+b}}$+$\frac{c}{\sqrt{1+c}}$ -- (*)

Does this means that I try to prove (*) by two cases:
1. assuming a>b and a>c
2. assuming a<b and a<c???

7. Mar 19, 2012

### Lily@pie

I'm not sure whether this way of proving it is okay:

I take a look at the function

f(x)=$\frac{x}{\sqrt{1+x}}$

f'(x)=$\frac{2+x}{2(1+x)^{\frac{3}{2}}}$
f'(x)>0 if and only if x≥-1.

Since I know that 0≤a≤b+c, f(x) is an increasing function in the interval [0,b+c]. Therefore, by choosing a=b+c, f(a)=$\frac{a}{\sqrt{1+a}}$=$\frac{b+c}{\sqrt{1+b+c}}$ is the maximum possible value.

Hence it is enough to show that $\frac{b+c}{\sqrt{1+b+c}}$$\leq$$\frac{b}{\sqrt{1+b}}$+$\frac{c}{\sqrt{1+c}}$

Since 1+b+c≥1+b and 1+b+c≥1+c,
$\frac{b+c}{\sqrt{1+b+c}}$=$\frac{b}{\sqrt{1+b+c}}$+$\frac{c}{\sqrt{1+b+c}}$$\leq$$\frac{b}{\sqrt{1+b}}$+$\frac{c}{\sqrt{1+c}}$

Therefore,
$\frac{a}{\sqrt{1+a}}$$\leq$$\frac{b}{\sqrt{1+b}}$+$\frac{c}{\sqrt{1+c}}$

It seems a bit dodgy

8. Mar 19, 2012

### Norwegian

Your solution is good. You use both the idea about f'>0 and the analysis of the denomintarors in one argument.

9. Mar 19, 2012

### Lily@pie

So is this method valid?