• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
  • Like us on facebook here

proofs question (1 Viewer)

cormglakes

New Member
Joined
Jan 4, 2021
Messages
18
Gender
Undisclosed
HSC
2021
20210322_210416.jpg
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.
 

cormglakes

New Member
Joined
Jan 4, 2021
Messages
18
Gender
Undisclosed
HSC
2021
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
Joined
Aug 2, 2019
Messages
1,806
Gender
Male
HSC
2021
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.
 
Last edited:

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,328
Gender
Male
HSC
N/A
Theorem 1: For all positive integers , , and ,



Proof: By contradiction

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.
 
Last edited:

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,328
Gender
Male
HSC
N/A
Theorem 1: For all positive integers , , and ,



Direct Proof:

 
Last edited:

CM_Tutor

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,328
Gender
Male
HSC
N/A
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
Joined
Mar 11, 2004
Messages
2,328
Gender
Male
HSC
N/A
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



Adding these equations, we get:

 

TheShy

Member
Joined
Nov 12, 2018
Messages
89
Gender
Male
HSC
2021
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
 
Joined
Jun 21, 2020
Messages
12
Gender
Male
HSC
2021
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

Member
Joined
Mar 4, 2018
Messages
85
Gender
Male
HSC
2020
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

Member
Joined
Mar 4, 2018
Messages
85
Gender
Male
HSC
2020
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
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top