Prove n^2<= ......

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November 16th 2009, 12:49 PM
chings

Prove n^2<= ......

Hi,

Much appreciate if anyone could help with this...

Prove

n^2 <= (a1+...+an).(1/a1+...+1/an)
November 16th 2009, 01:21 PM
Raoh

hi(Hi)
$(a_{1}+....+a_{n})(\frac{1}{a_{1}}+...+\frac{1}{a_ {n}})=\sum_{k=1}^{n}a_{k}\sum_{k=1}^{n}\frac{1}{a_ {k}}=\sum_{k=1}^{n}1=n$
November 16th 2009, 01:28 PM
Plato

Quote:

Originally Posted by Raoh

hi(Hi)
$(a_{1}+....+a_{n})(\frac{1}{a_{1}}+...+\frac{1}{a_ {n}})=\sum_{k=1}^{n}a_{k}\sum_{k=1}^{n}\frac{1}{a_ {k}}=\sum_{k=1}^{n}1=n$

OH!
$(1+2)\left(1+\frac{1}{2}\right)=\frac{9}{2}$ .
November 16th 2009, 01:30 PM
Raoh

Quote:

Originally Posted by Plato

OH!
$(1+2)\left(1+\frac{1}{2}\right)=\frac{9}{2}$ .

hhh ! where i went wrong ?
thanks (i think i'm dizzy (Worried) )
November 16th 2009, 01:48 PM
tonio

Quote:

Originally Posted by chings

Hi,

Much appreciate if anyone could help with this...

Prove

n^2 <= (a1+...+an).(1/a1+...+1/an)

It must be $a_i>0\,\,\,\forall\,\,i$ , otherwise $2^2\nleq (1+(-1))\left(\frac{1}{1}+\frac{-1}{1}\right)=0$

Now, do you know the Means Inequalities?:

$\frac{a_1+...+a_n}{n}\,\,\geq \,\,\sqrt[n]{a_1\cdot ...\cdot a_n}\,\,\geq\,\, \frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}}$

Well, taking the two extremes in the above inequalities gives you what you want. About the proof of the inequalities (well, THE inequality, since only the left one is needed: the right one follows from this taking the inverse of the elements) you can search inside MHF and look for a solution. I sent one just yesterday, in another thread)

Tonio
November 16th 2009, 01:53 PM
Defunkt

Quote:

Originally Posted by Raoh

hhh ! where i went wrong ?
thanks (i think i'm dizzy (Worried) )

Well, what you said is just not right :x

$\sum_k a_k \cdot \sum_i b_i = a_1(\sum_i b_i) + a_2(\sum_i b_i) + ... \neq a_1b_1 + a_2b_2 + ...$
November 16th 2009, 01:55 PM
Raoh

Quote:

Originally Posted by Defunkt

Well, what you said is just not right :x

$\sum_k a_k \cdot \sum_i b_i = a_1(\sum_i b_i) + a_2(\sum_i b_i) + ... \neq a_1b_1 + a_2b_2 + ...$

thank you.(Wink)
November 16th 2009, 02:09 PM
chings

Quote:

Originally Posted by tonio

It must be $a_i>0\,\,\,\forall\,\,i$ , otherwise $2^2\nleq (1+(-1))\left(\frac{1}{1}+\frac{-1}{1}\right)=0$

Now, do you know the Means Inequalities?:

$\frac{a_1+...+a_n}{n}\,\,\geq \,\,\sqrt[n]{a_1\cdot ...\cdot a_n}\,\,\geq\,\, \frac{n}{\frac{1}{a_1}+...+\frac{1}{a_n}}$

Well, taking the two extremes in the above inequalities gives you what you want. About the proof of the inequalities (well, THE inequality, since only the left one is needed: the right one follows from this taking the inverse of the elements) you can search inside MHF and look for a solution. I sent one just yesterday, in another thread)

Tonio

Hi all,

Thanks for the responses. Tonio: don't know much about the mean inequalities but yes indeed, ai are positive real numbers. and begging your pardon but what is MHF? I'll search for the post for the proof for the above.
November 16th 2009, 02:56 PM
Drexel28

Quote:

Originally Posted by chings

... pardon but what is MHF? .

I'll give you a hint, it's big, blue, and you just logged onto it.
November 16th 2009, 05:45 PM
chings

zomg noobed.

thanks again.
November 17th 2009, 05:47 AM
chings

Hi Tonio,

still can't seem to find the page. Could you kindly assist please thank you.
November 17th 2009, 05:50 AM
Raoh

Quote:

Originally Posted by chings

Hi Tonio,

still can't seem to find the page. Could you kindly assist please thank you.

you mean MHF ?(Happy)
November 17th 2009, 05:52 AM
tonio

Quote:

Originally Posted by chings

Hi Tonio,

still can't seem to find the page. Could you kindly assist please thank you.

It's here: http://www.mathhelpforum.com/math-he...-exercise.html

The second post there is mine and where I prove the inequality.

Tonio