Olympiad problem

[loebas]

a+b+c=1; a,b,c ∈ R+

Find the max of:


My guess would be {2,2,-3} but cannot prove it, can somebody help?

 

[Deleted member 4993]

a+b+c=1; a,b,c ∈ R+

Find the max of:
View attachment 31655

My guess would be {2,2,-3} but cannot prove it, can somebody help?

This is a problem from COMPETITION. Are you allowed to seek external help?

a+b+c=1; a,b,c ∈ R+

Find the max of:

\(\displaystyle \frac{1}{a^2-4a+9} + \frac{1}{b^2-4b+9} + \frac{1}{c^2-4c+9} \)

Why would you guess (2, 2, -3)? Please state your reason for this guess?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[loebas]

It's an old olympiad from Korea in 2011, so I guess yes everybody is allowed to look for external help.
There are 3 terms, the maximum of each term is 1/5 (when a, b, or c =2), obtained by a simple derivation.
As a or b or c tends to infinite, every terms tends to 0. My reasoning is intuitive : a+b+c = 1, and in order to maximize the sum of the terms, I would be as close possible as the value 2. By choosing (2/2/-3), or any combination of these, I get 13/30 as a maximum, which is higher than other combinations i tried by getting further (eg. 3/1/-3, etc.). I am just stuck on how i should begin to prove this.

 

[loebas]

I realize that a,b,c should be positive so forget my attempt

 

[BigBeachBanana]

It's an old olympiad from Korea in 2011, so I guess yes everybody is allowed to look for external help.
There are 3 terms, the maximum of each term is 1/5 (when a, b, or c =2), obtained by a simple derivation.
As a or b or c tends to infinite, every terms tends to 0. My reasoning is intuitive : a+b+c = 1, and in order to maximize the sum of the terms, I would be as close possible as the value 2. By choosing (2/2/-3), or any combination of these, I get 13/30 as a maximum, which is higher than other combinations i tried by getting further (eg. 3/1/-3, etc.). I am just stuck on how i should begin to prove this.

Isn't [imath]\R+[/imath] means positive real numbers only? c=-3 violates that.
Since a+b+c=1 and the fractions are "symmetrical", I think a good guess would've been a=b=c=1/3.