Few suggestions about cauchy inequality

[topengonzo]

As I can see from the formula of cauchy inequality:
(a1^2+a2^2+...+an^2)^1/2 . (b1^2+b2^2+...+bn)^1/2 >= a1b1+a2b2 + ... + anbn

Can I conclude from the above formula that:
(a1+a2+...+an)^1/2 . (b1+b2+...+bn)^1/2 >= (a1b1)^1/2 + (a2b2)^1/2 +...+ (anbn)^1/2
by setting a1,...,an = p1^2,...pn^2 and b1,...,bn = q1^2,...qn^2

Also in my lectures they mentioned this formula
(a1+a2...+an)(b1+b2+...bn) <= n(a1b1+a2b2+...+anbn)
but they didnt give a proof for it. Anyone can find me proof for it as I don't know how to prove it.

 

[awkward]

As I can see from the formula of cauchy inequality:
(a1^2+a2^2+...+an^2)^1/2 . (b1^2+b2^2+...+bn)^1/2 >= a1b1+a2b2 + ... + anbn

Can I conclude from the above formula that:
(a1+a2+...+an)^1/2 . (b1+b2+...+bn)^1/2 >= (a1b1)^1/2 + (a2b2)^1/2 +...+ (anbn)^1/2
by setting a1,...,an = p1^2,...pn^2 and b1,...,bn = q1^2,...qn^2

True, provided the a's and b's are non-negative.

Also in my lectures they mentioned this formula
(a1+a2...+an)(b1+b2+...bn) <= n(a1b1+a2b2+...+anbn)
but they didnt give a proof for it. Anyone can find me proof for it as I don't know how to prove it.

False. Consider, for example, the case n=2, a1 = b2 = 0, a2 = b1 = 1.

 

[topengonzo]

True, provided the a's and b's are non-negative.False. Consider, for example, the case n=2, a1 = b2 = 0, a2 = b1 = 1.

Your right. Rereading the lecture, they mentioned that that a1,..,an and b1,...,bn are in descending order and they are all positive.

 

[topengonzo]

How to prove it?

 

[awkward]

Have you tried induction?

 

[Norwegian]

The last in inequality is named after Chebyshev. Lots of different spellings, but I am sure you can google it. The sequences must be sorted the way, and it is not requiered that the terms are positive.