1. ## Prove n^2<= ......

Hi,

Much appreciate if anyone could help with this...

Prove

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

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

3. Originally Posted by Raoh
hi
$\displaystyle (a_{1}+....+a_{n})(\frac{1}{a_{1}}+...+\frac{1}{a_ {n}})=\sum_{k=1}^{n}a_{k}\sum_{k=1}^{n}\frac{1}{a_ {k}}=\sum_{k=1}^{n}1=n$
OH!
$\displaystyle (1+2)\left(1+\frac{1}{2}\right)=\frac{9}{2}$.

4. Originally Posted by Plato
OH!
$\displaystyle (1+2)\left(1+\frac{1}{2}\right)=\frac{9}{2}$.
hhh ! where i went wrong ?
thanks (i think i'm dizzy )

5. 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 $\displaystyle a_i>0\,\,\,\forall\,\,i$, 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

6. Originally Posted by Raoh
hhh ! where i went wrong ?
thanks (i think i'm dizzy )
Well, what you said is just not right :x

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

7. Originally Posted by Defunkt
Well, what you said is just not right :x

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

8. Originally Posted by tonio
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

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.

9. 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.

10. zomg noobed.

thanks again.

11. Hi Tonio,

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

12. Originally Posted by chings
Hi Tonio,

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

13. 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