proofs question (1 Viewer)

cormglakes

New Member
Also another proofs question,
how the bloody hell do u prove :

a/(b+c) + b/(a+c) + c/(a+b) > 3/2

also does anyone have proofs tips? coz i rlly suck at proofs.

idkkdi

Well-Known Member
Also another proofs question,
how the bloody hell do u prove :

a/(b+c) + b/(a+c) + c/(a+b) > 3/2

also does anyone have proofs tips? coz i rlly suck at proofs.
View attachment 30433
so i can do question A by doing LHS- RHS and then doing 2 cases where a>b and a<b so show LHS - RHS >0 ; is there and alternative solution?

and for B and C, I have absolutely no clue.; for B i think you have to add multiple inequalities but i am not sure what inequalities.
These one's are pretty hard. I've done them before but I seriously can't be bothered to write it.
A few ways for both.
1. Amgm with two or three terms can get the job done. Imaginary terms should be created. Referring to part C.
2. Am-gm with three terms of a two term method also exists. Start from something else to get to this answer, backwards reasoning may help but experience likely needed.

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CM_Tutor

Moderator
Moderator
Theorem 1: For all positive integers , , and ,

Assume the theorem is false. Then there exist positive integers , , and , such that

This is a contradiction as is an integer and so its square cannot be negative.

Every statement following the assumption is true, provided the assumption is true - and yet, the result is a statement that is false. This is only possible if the assumption itself is false, and thus the theorem must be true.

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CM_Tutor

Moderator
Moderator
Theorem 1: For all positive integers , , and ,

Direct Proof:

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• B1andB2

CM_Tutor

Moderator
Moderator
Theorem 2: For all positive integers , , and ,

Direct Proof:

From Theorem 1, we know that

And by changing the dummy variables, we also know that

and that

Adding (1) + (2) + (3) gives:

Notice that we have, on the RHS of (*), three pairs of terms all of the form

where and are positive integers. This can be simplified by recognising that each of these pairs of terms has a minimum value of 2:

Applying this result to (*) gives:

CM_Tutor

Moderator
Moderator
Theorem 3: For all positive integers , , and ,

Direct Proof:

In the direct proof of Theorem 1, it was established that

from which it can be shown (by division by ) that:

Using the same approach with and , we find that:

and with and , we get

TheShy

Active Member
a/(b+c) + b/(a+c) + c/(a+b) > 3/2

You can consider doing this by considering the fact that: (a+b+c)(1/a + 1/b + 1/c)>=9. Of course you have to prove that first, but its very simple, just expand and use the fact that a+1/a >=2.
Anyways, considering that, let a=b+c, b=a+c, c=a+b. Sub those in, then expand, and you get the desired result

slightly_above_average

New Member
for part a)
Expand right hand side
bring terms to the left
factorise by (a-b)
so u get (a^2-b^2)(a-b)
so youll get (a+b)(a-b)^2 which should right

s97127

Active Member
a^3+b^3+c^3 - 3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (a+b+c) * 1/2 * [(a-b)^2 + (b-c)^2 + (c-a)^2] >= 0

This one is stronger as we don't need a, b and c are positive numbers. We only need a+b+c >= 0

s97127

Active Member
a/(b+c) + b/(a+c) + c/(a+b) > 3/2

let x = a+b, y = a+c and z = a+b
the problem will become:
1/2 * [(y+z-x)/x + (x+z-y)/y + (x+y-z)/z) >= 3/2
(y/x + x/y) + (z/x + x/z) + (y/z + z/y) >= 6
this is easy as y/x + x/y >= 2