Contradiction, i think (1 Viewer)

[Logix]

Ok, this is from the cambridge book from harder 3 unit chapter. question is:

Show that:

1/a + 1/b + 1/c >= 9

Condition is if a+b+c=1.

(a+b+c)(1/a+1/b+1/c) >= 9(a+b+c)
>= 9

when u multiply both sides by (a+b+c),
LHS = 1+1+1+ (b/a+a/b) + (b/c+c/b) + (a/c+c/a)
>= 3+2+2+2
>= 9
This is the proof that i could get using the arithmetic - geometric identity.

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This is the other method i tried to use that failed.

instead of multilying both sides by (a+b+c), I added (a+b+c) to both sides.
So i had to prove that (a+b+c) + (1/a+1/b+1/c) >= 9+(a+b+c)
>= 10

LHS = a+b+c+1/a+1/b+1/c
= (a+1/a) + (b+1/b) + (c+1/c)
>= 2+2+2
>= 6 Using the arithmetic - geometric identity again.
But that doesnt prove it is >=10.....
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am i doin something really stupid or is the question wrong?

Thx

 

[Estel]

Your method is rather flawed:
(a+1/a) + (b+1/b) + (c+1/c) is most certainly not 6.

 

[Xayma]

This is the other method i tried to use that failed.

instead of multilying both sides by (a+b+c), I added (a+b+c) to both sides.
So i had to prove that (a+b+c) + (1/a+1/b+1/c) >= 9+(a+b+c)
>= 10

LHS = a+b+c+1/a+1/b+1/c
= (a+1/a) + (b+1/b) + (c+1/c)
= 2+2+2
= 6 Using the arithmetic - geometric identity again.
But 6 isnt >=10.....
--------------------------------------------------------------------------------------------
am i doin something really stupid or is the question wrong?

Thx

LHS=(a+1/a)+(b+1/b)+(c+1/c)
=(a²+1)/a+(b²+1)/b+(c²+1)/c

I don't see how you got 2's.

 

[Logix]

i forgot to write it is greater or equal to 6 (a+1/a+b+1/b+c+1/c >=6)

this is from (a+1/a)^2>=4
therefore (a+1/a)>=2

 

[Logix]

but how does expressing it like that get any closer to a numerical answer?

 

[Logix]

no. i got it from the arithmetic-geometric inequality.
lemme show u how...

(a+b)^2>=4ab (this is the inequality)
Leave a as is....Sub b=1/a,
therefore (a+1/a)^2>=4(a)(1/a)
(a+1/a)^2>=4
(a+1/a)>=2
Same for b and c,
So (a+1/a)+(b+1/b)+(c+1/c)>=2+2+2>=6!

 

[Estel]

a + 1/a = 2 if and only if a = 1
b + 1/b = 2 if and only if b = 1
c + 1/c = 2 if and only if c = 1

but a + b + c = 1
hence a+1/a + b+1/b + c + 1/c is NOT equal to 6.
[keeping in mind that a, b and c are all positive, and that a + 1/a is larger than 2 if a <>1, etc]

You cannot add inequalities if you do not know what you are doing.

 

[dawso]

yeah. i tried it wit various algebraic methods and got variuos answers, ie- =3, 6
but if u think about it logically and put numbers in in place of a,b,c then u get all answers greater /equal to 9, (=9 if a,b,c =1/3) so i dunno how 2 prove it but the question aint wrong
-dawso

 

[Logix]

firstly to estel..................it was a typo. i meant >=2!!

and to dawso..... wat methods did u use? i racked my brains to think of those 2....lol
u must b a genius if u can think of other legitimate methods

 

[Estel]

"i meant >=2!!"
You're right there; but the leap of logic you used is why you're wrong.
Adding 3 inequalities which are >=2 doesn't result in >=6! (as prove above)

There are other methods of doing your inequality; a lot involve non-Yr12 material.

 

[Logix]

wats ur problem today? i made a little typo, and ur still goin on bout it! ease up man! apart from forgetting to add >=, how else have i been incompetent in this area?

well sorrry, if i cannot add inequalities!

 

[Logix]

apart from that typo, can any1 see a PROPER error?

 

[ngai]

hmm...a contradiction, u think?
so the whole world of maths falls apart?
its amazing how many ppl dont realise whats going on in these situations...
theres nothing wrong...that crap is >= 6
in fact its >= 9, as proven in ur 1st method
just like the number 50
50 > 6, and 50 > 9
but u cant prove 50 > 9 using 50 > 6

 

[Logix]

of course maths isnt real anyway! thats y we have complex

but do u's see anything wrong? i bet its something real obvious staring us in the face. however, i cant see it

 

[ngai]

hehe look carefully
and u'll see..

stuff like this heheh
invisible text is fun : D

 

[Rorix]

yeah, careful inspection of the 3rd line and onwards should give you the answer

ngai, are you going tomorrow? I should ask on MSN but this is cooler

 

[Archman]

bah, as i said in the other thread, a stronger result implies a weaker one, but not the other way round.

 

[Logix]

i still dont see it

 

[McLake]

My sillyness is contageous ...

 

[Xayma]

What you did was the following:
(a+1/a)²+(b+1/b)²+(c+1/c)²&ge;6

This is true, in fact due to a+b+c=1, it is in fact larger then this.

However you then took it as equal to 6.

Which it isn't.

Another example:

6&ge;4
6&ge;5

What you did was take the first line and its lowest value and tried to position it in the second ie
4&ge;5. Which is false.

What archman is saying is that 6&ge;5 is the stronger case if this is true then 6&ge;4, whereas 6&ge;4 being the weaker case does not indicate anything about 6&ge;5 if it is true.