An inequality question

[nfornick]

Show that

a/c + b/a + c/b > a+b+c, for a, b, c > 0 and a*b*c<1

[BigMoosie]

[math]\frac{a}{c} + \frac{b}{a} + \frac{c}{b} > a+b+c[/math]

 

[math]a + b + c > (a+b+c)(abc)[/math]

 

[math]0 > (a+b+c)(abc - 1)[/math]

[timo]

That´s not correct. The 2nd line would be: a²b + b²c + c²a > (a+b+c)(abc)

[albertlee]

extending from Atheist,

 

a²b + b²c + c²a > (a+b+c)(abc)

==> a²b + b²c + c²a > a²bc+ab²c+abc²

 

therefore:

 

0 > (c-1)a²b+(a-1)b²c+(b-1)ac²

[Mobius]

(c-1)a2b+(a-1)b2c+(b-1)ac2

 

And how are you proving that this is a negative number?

 

I don't seem to be ble to solve this inequality mathematically BUT I can solve it logically.

 

In order to satisfy a*b*c<0 where they are all positive values then at least ONE of them must be a fraction and the other two numbers cannot multiply together to be greater than the denominator.

 

e.g. 1/20, 3, 5... When multiplied this gives 3/4.

 

Now:-

a/c + b/a + c/b > a+b+c

 

On the left hand side a, b and c are all denominators, one of them will be a fraction (say a) and as it is on the bottom line it will be inverted and will be large. It will also be larger then the other two numbers added together (b + c) due to my logic above (a*b*c<0). i.e. if it is larger than the product it will be larger than the sum. So the right hand side will be b + c plus the fraction which will be smaller than the left had side.

 

Of course if a, b and c are all fractions then this inequality is obvious as when fractions are divided the answer is always greater (or equal) then either of the two fractions to begin with.

 

this is not very mathematical and I'm sure if there is a flaw in my logic it will be pointed out.... It is a logical proof however (I did try for a mathematical one!).

[albertlee]

let me rephrase...

[Mobius]

Your last equation 0 > (c-1)a^2b+(a-1)b^2c+(b-1)ac^2, would make sense if a, b and c were ALL fractions, but as I pointed out in my last post only 1 of them has to be a fraction (according to the original conditions).

 

I think my proof is fine but it may have a flaw... Awaiting confirmation!!!!

[BigMoosie]

That´s not correct. The 2nd line would be: a²b + b²c + c²a > (a+b+c)(abc)

 

oops, stupid me

[nfornick]

Re: Mobius

 

a*b*c<1 but not a*b*c<0

 

"at least ONE of them must be a fraction"

means one or two may be a fraction

 

"one of them will be a fraction (say a) and as it is on the bottom line it will be inverted and will be large. It will also be larger then the other two numbers added together (b + c)"

a*b*c<1 does not imply

1/a>b+c

say a=1/3, b=1, c=2

[Mobius]

"at least ONE of them must be a fraction"

means one or two may be a fraction

 

Doesn't matter of 2 of them are fractions, this just strengtens the case.

 

1/a>b+c

say a=1/3, b=1, c=2

 

That is true, in this case 1/a = b + c

However there are two more numbers to be added to that 1/a (or b/a) as it would be.

Now I did say it would be greater but as you showed it could be equal, however as there are 2 more numbers to be added to it, it will always be greater...

[dttom]

L.H.S.

a/c + b/a + c/b

=[(a^2)b+(b^2)c+(c^2)a]/abc

let x as abc which is smaller than 1

=(a^2b+b^2c+c^2a)/x

L.H.S.*x=a^2b+b^2c+c^2a

R.H.S*x=(a+b+c)abc

=(a^2)bc+a(b^2)c+ab(c^2)

=(c-1)(a^2b)+(a-1)(b^2c)+(b-1)(c^2a)+a^2b+b^2c+c^2a

let LHS*x as r and RHS*x as w,

r-r=0

w-r=(c-1)(a^2b)+(a-1)(b^2c)+(b-1)(c^2a)

0>abc(a+b+c)-(a^2b+b^2c+c^2a)