Hi,

Much appreciate if anyone could help with this...

Prove

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

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- Nov 16th 2009, 12:49 PMchingsProve n^2<= ......
Hi,

Much appreciate if anyone could help with this...

Prove

n^2 <= (a1+...+an).(1/a1+...+1/an) - Nov 16th 2009, 01:21 PMRaoh
hi(Hi)

$\displaystyle (a_{1}+....+a_{n})(\frac{1}{a_{1}}+...+\frac{1}{a_ {n}})=\sum_{k=1}^{n}a_{k}\sum_{k=1}^{n}\frac{1}{a_ {k}}=\sum_{k=1}^{n}1=n$ - Nov 16th 2009, 01:28 PMPlato
- Nov 16th 2009, 01:30 PMRaoh
- Nov 16th 2009, 01:48 PMtonio

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

Now, do you know the Means Inequalities?:

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

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 - Nov 16th 2009, 01:53 PMDefunkt
- Nov 16th 2009, 01:55 PMRaoh
- Nov 16th 2009, 02:09 PMchings
- Nov 16th 2009, 02:56 PMDrexel28
- Nov 16th 2009, 05:45 PMchings
zomg noobed.

thanks again. - Nov 17th 2009, 05:47 AMchings
Hi Tonio,

still can't seem to find the page. Could you kindly assist please thank you. - Nov 17th 2009, 05:50 AMRaoh
- Nov 17th 2009, 05:52 AMtonio

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

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

Tonio