Prove that if :
( )
Suppose one of is 0. Then the other two can’t be 0 if the condition is to hold. Say, we have . Then and the inequality becomes
The LHS, as a quadratic expression in , has negative discriminant and so the inequality is true.
So now we’ll assume . Let . Then and the inequality becomes
Since are positive and , we can make the following substitutions:
where . Then observe that
Hence
So it comes down to proving this inequality:
hmmm... don’t know why I was heavily neg repped on this question =\ but yer .. anyways...
^^ top was interesting
After about 2 pages of working trying to find similarities between the two I got this...
LHS
8(a+b+c)^2
8a^2 + 16ab + 16ac + 8b^2 + 16bc + 8c^2
8(a^2 + b^2 + c^2) + 16(ac + ab + cd)
8(a^2 + b^2 + c^2) + 16(3)
8(a^2 + b^2 + c^2) + 48
8(a^2 + b^2 + c^2 + 6)
RHS
9(a+b)(a+c)(b+c)
9a^2b + 9a^2c + 9ab^2 + 18abc + 9ac^2 + 9b^2c + 9bc^2
9a(ab + ac) + 9b(ab + bc) + 9c(ac + bc) +18(abc)
9a(3-bc) + 9b(3-ac) + 9c(3-ab) +18abc
27a – 9abc + 27b – 9abc + 27c -9abc + 18abc
27(a + b + c) – 9abc
9(3(a + b + c) - abc)
So we can say….
8(a^2 + b^2 + c^2 + 6) => 9(3(a + b + c) - abc)
when ab + cb + bc = 3…..
I also noticed a + b + c appears In the RHS which also appears in the LHS
So we can let a + b + c = d
And say:
8d^2 => 27d – 9abc
still doesn’t help unless we know a b or c = 0…