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Suppose sum of the 7 positive numbers is 21. What is the minimum possible value of the averages of square of these numbers-- a. 63 b. 21 c. 9 d. 7 Can anyone help me please...

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In geometric terms, what is the point on the hyperplane $\sum x_i=21$ that is closest to the origin? –  lhf Jun 20 '13 at 14:40
Alternative to lhf's method, do you know what AM-GM is? –  Calvin Lin Jun 20 '13 at 14:41
Your posts seem to have the uncanny tendency to be void of any and all demonstrated effort on your part. All effort shown is in the answers you receive. That is going to hurt you in the long run. By now, I would think you would have "caught on" to the expectations of the site. –  amWhy Jun 20 '13 at 14:53

2 Answers 2

up vote 2 down vote accepted

Using Cauchy–Schwarz inequality, putting $b_i=1$ for $1\le i\le n,$

$$n\cdot \sum_{1\le i\le n}a_i^2\ge (\sum_{1\le i\le n}a_i)^2$$

So, $$7\cdot \sum_{1\le i\le 7}a_i^2\ge (\sum_{1\le i\le 7}a_i)^2=21^2$$

$$\implies \frac{\sum_{1\le i\le 7}a_i^2}7\ge 9$$

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@ lab bhattacharjee I amn't getting the last line it should be 21*3 instead of 9 ? –  Prasanta Jun 20 '13 at 15:11
@Prasanta. $$\frac{21^2}{7\cdot 7}=9,$$ right? –  lab bhattacharjee Jun 20 '13 at 15:42
okkk... I got it... –  Prasanta Jun 20 '13 at 16:07

The "rule of thumb" in such problems is that extrema correspond to identical parameters. I.e., all 7 numbers are the same (3) and the average square is 9.

The actual proof follows from the root mean squared inequality.

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