inequality (1 Viewer)

[Monsterman]

I proved x^2+y^2 >= 2xy
how would i use my result to deduce that (a+b)(b+c)(c+a)>=8abc?

 

[Diaz]

So multiplying these together gives:



As required.

 

[nonsenseTM]

I expanded the LHS and got a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc, so -2abc on both sides, so we need to prove a^2b+ab^2+a^2c+ac^2+b^2c+bc^2 is greater than or equal to 6abc.

using the a^2+b^2 >= 2ab, a^2c+b^2c >= 2abc
similarly the other two groups are >= 2abc
so the six terms are >= 6abc

 

[nonsenseTM]

So multiplying these together gives:



As required.

much smarter than my way forgot the 2

 

[Diaz]

Erp. Tripped up. Thanks.
Just realised - my way does assume positive a,b,c, so your way's probably more rigorous.

 

[Monsterman]

^
Thanks alot.. and the question said the numbers are non-negative.. so yea..

Another question which continues from the continued question.......
(b+c)(c+a)(a+b)>8abc
let x,y,z be unequal positive numbers such that x+y>z, y+z>x, z+x>y..
Use the (b+c)(c+a)(a+b)>8abc result to prove that (x+y-z)(y+z-x)(z+x-y)< xyz

 

[untouchablecuz]

^
Thanks alot.. and the question said the numbers are non-negative.. so yea..

Another question which continues from the continued question.......
(b+c)(c+a)(a+b)>8abc
let x,y,z be unequal positive numbers such that x+y>z, y+z>x, z+x>y..
Use the (b+c)(c+a)(a+b)>8abc result to prove that (x+y-z)(y+z-x)(z+x-y)< xyz

 

[Monsterman]

ty... i can never seem to figure out what to do with these questions..

 

[shaon0]

ty... i can never seem to figure out what to do with these questions..

lol, is this for the upcoming MATH1151 calc test?

 

[Monsterman]

lol, is this for the upcoming MATH1151 calc test?

awesome guess

 

[gurmies]

awesome guess

I thought so too

 

[shaon0]

Bad luck, there was no inequality questions for us today.

 

[Monsterman]

Bad luck, there was no inequality questions for us today.

better safe than sorry..