Inequalities q (2 Viewers)

[iStudent]

Can we do this:
RTP 1/a^2 + 1/b^2 + 1/c^2 >= 27 (cambridge q. ex 8.1 q8)
given a + b + c = 1 and a>0, b>0, c>0

Consider:
(a^2+b^2+c^2)(1/a^2 + 1/b^2 + 1/c^2) >= -54(ab+bc+ac) which is true (I think ..) since RHS is negative whereas LHS is positive
>= 27(-2(ab+bc+ac)
>= 27(a^2+b^2+c^2 - (a+b+c)^2))
>=27(a^2+b^2+c^2) since a+b+c = 1
so
(a^2+b^2+c^2)(1/a^2 + 1/b^2 + 1/c^2) >= 27(a^2+b^2+c^2)
hence,
1/a^2 + 1/b^2 + 1/c^2 >= 27

I know there is something very wrong about this but I can't put my finger on it. Can someone explain please?
Is there a particular inequality rule I don't know about that you can't use the expansion (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ac) in some cases?

 

[seanieg89]

27(a^2+b^2+c^2 - (a+b+c)^2)) >= 27(a^2+b^2+c^2) since a+b+c = 1

Can you explain this?

 

[iStudent]

It was part of the assumption in the question

 

[seanieg89]

It was part of the assumption in the question

I know a+b+c = 1 is an assumption, but why should this imply the inequality I quoted?

 

[iStudent]

The full question was:

If a>0, b>0, c>0 and a + b + c = 1, show that:
a. 1/a + 1/b + 1/c >= 9
and
1/a^2 + 1/b^2 + 1/c^2 >= 27
I did the first part already

 

[iStudent]

I see.

 

[iStudent]

now, 27(a^2+b^2+c^2) = = 27(a^2+b^2+c^2 - (a+b+c)^2)) = 27(-2(ab+bc+ac) since a + b + c = 1
but
(a^2+b^2+c^2)(1/a^2 + 1/b^2 + 1/c^2) >= -54(ab+bc+ac) as RHS is negative
hence, (a^2+b^2+c^2)(1/a^2 + 1/b^2 + 1/c^2) >= 27(a^2+b^2+c^2)

does that change anything?

 

[seanieg89]

27(a^2+b^2+c^2 - (a+b+c)^2))=27(a^2+b^2+c^2) since a+b+c = 1

?

 

[iStudent]

Ok nvm i see where i went wrong

 

[iStudent]

Yea I mistook it as a + b +c = 0.
>.<
My brain screwed up after doing too many inequalities questions...
Thanks.