How to "show" something in math

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G

GoldenBear

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Mar 2, 2000
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This homework question asks:

Let a, b, c, be nonnegative numbers. Show that

if a <= b + c, then a/(1+a) <= b/(1+b) + c/(c+c)

I can just plug in numbers and do it that way..but that seems to be a bit too easy, is there a SPECIFIC way I'm supposed to show this?

And also, is there a difference between showing something, and proving something?
 
G

GoldenBear

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<< why do we have to do your homework for you? >>

Because you're all very intelligent and giving people.
 
G

GoldenBear

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Mar 2, 2000
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<< 1+1=2 >>

I didn't realize exactly how far your intellect ability went, you surprise me sometimes.
 
D

darkshadow1

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Nov 2, 2000
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Usually for problems like that you can't use specific numbers...were there like rules or such that you guys went over recently? You may have to use a combination of like transitive, additive, etc...this smells sorta like discrete math...not quite icky enough though. :p
 
K

Kelvrick

Lifer
Feb 14, 2001
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I graduated last week. You expectin me to think? Umm... Damn, he already sayd 1+1=2. Thats as far as I can go now... I'll get back to you after some more research.
 
G

GoldenBear

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<< Show? It's SHEW damnit, SHEW!


I doubt anyone will know what I'm talkin about, but hey.. >>

Here's my finger here's my thumb, here's a fist you better run.
 
P

Pretender

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Mar 14, 2000
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alright, let me see how to help

1 = a/a
a/(1+a) = 1 + a
b/(1+b) = 1 + b
c/(1+c) = 1 + c

so we need to show that
if
a <= b + c,
then
1+a <= 1+b + 1+c

if it isn't obvious, I'll keep going

1+a <= b + c + 2
a-1 <= b + c
if (a <= b + c), then (a-1 <= b + c) also

 
H

Haircut

Platinum Member
Apr 23, 2000
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Pretender, a/(1+a) does not equal 1+a
Take the number 3 as an example, 3/(1+3)=3/4, not 4
 
Belegost

Belegost

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Feb 20, 2001
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Ummm how exactly did you get a/(1+a) = 1+a?

Also I'm going to assume GB made a typo in the original post and meant c/(1+c) instead of c(c+c).
I'll be back when I've gotten an answer to give some hints on where to go. (assuming I'm able to get an answer)
 
H

Haircut

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Apr 23, 2000
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I think I've got it now.
a/(1+a)<=b+c/(1+b+c) from the fact that a<=b+c
now b/(1+b)+c/(1+c)=b+c+2bc/(1+b+c+bc) simple addition of fractions
we can separate this out into b+c/(1+b+c+bc) + 2bc/bc
= (b+c/(1+b+c+bc)) + 2
b+c/(1+b+c) <= 1
so (b+c/(1+b+c+bc)) +2 >= b+c/(1+b+c) as b+c/(1+b+c+bc) >= 0
so we have now that
a/(1+a) <= b+c/(1+b+c) <= (b+c/(1+b+c+bc)) + 2
so a/(1+a) <= (b+c/(1+b+c+bc)) + 2
a/(1+a)<=b/(1+b)+c/(1+c)

I think that's right but I've not bothered to check properly yet!
<edit> just noticed it's wrong. I can't be bothered to fix it as i've got to go out now, but I think this is along the lines of how it should go </edit>
 
L

lucidguy

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Apr 24, 2001
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<< b/(1+b)+c/(1+c)=b+c+2bc/(1+b+c+bc) simple addition of fractions
we can separate this out into b+c/(1+b+c+bc) + 2bc/bc
 >>



Nope.
 
P

Pretender

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Mar 14, 2000
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i forgot that you can't do the same things with numerators and denominators

i was thinking

(a+1)/a = 1 + 1/a, but that sorta expansion doesn't work for denoms.

a/(1+a) -> [(a + 1) - 1]/(a+1) -> 1 - 1/(a+1)
dunno if that helps
 
Belegost

Belegost

Golden Member
Feb 20, 2001
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GRRR NOT ERROR!
I thought i made a mistake in my logic, but I don't think I did.
Anyhow, what I tried:
Add fractions:
a/(a+1) <= (b+2bc+c)/(b+bc+c+1)
cross multiply:
ab+abc+ac+a <= ab+b+2abc+2bc+ac+c
Subtract out like terms:
a <= b+c+abc+2bc
Which must be true if a <= b+c since,
a <= b+c <= b+c+abc+2bc if, a,b,c are non-negative real numbers.

I hope I did my math correctly, someone correct me if I didn't.
 
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