Possible average square value suppose the sum of seven Positive number is 21. what is the minimum possible value of the average of the square of these number?
 A: Assume $\sum_{i=1}^7 x_i = 21$. From Jensen's inequality with $\varphi(x) = x^2$ and all weights equal to $1$, we get:
$$\varphi \left(\frac {\sum_{i=1}^7 x_i} 7 \right) \leq \frac {\sum_{i=1}^7 \varphi(x_i)} 7$$
Plugging in what we know, on the left hand side we have simply $\varphi(3) = 9$, and on the right side we have the average of the squares. So the answer is at least 9.
Easy exercise: Prove it's exactly 9; that is, find 7 numbers that sum to 21 and have 9 as their average square.
A: $$\sum_{i=1}^n x_i = 21 \tag{1}$$
Minimizing the average of the seven sqares is the same as minimizing the sum of the squares. Both values differ by the constant facor 7. 
So we proceed by minizing the sum of the suqares, The latter expression is simpler so we prozeed by minimizing
$$\sum_{i=1}^7x_i^2 \tag{2}$$.
We can  write 
$$
\begin{eqnarray}
\tag{3} \\
x_1=3+d_1 \\
x_2=3+d_2 \\
\cdots    \\
x_7=3+d_7 \\
\end{eqnarray}
$$
Substituting  $(3)$ in $(1)$ gives
$$\sum_{i=1}^n d_i = 0 \tag{4}$$
So we have 
$$
\begin{eqnarray}
\sum_{i=1}^7 x_i^2  = \\
\sum_{i=1}^7 (3+d_i)^2 = \\
\sum_{i=1}^7 (9 + 6 d_i +d_i^2) = \\
63 + 6 \sum_{i=1}^7d_i + \sum_{i=1}^7 d_i^2 = \\
63 + \sum_{i=1}^7 d_i^2 \tag{5}
\end{eqnarray}
$$
The sum of $d_i$ vanishes because of $(4)$  $(5)$ is  $\gt 63$ if some $d_i \gt 0$. It is equal to 63 and therfore 
minimal if all $d_i=0$.
This is equivalent to all $x_i=3$.
A: Let's assume we have a 7-tuple in which some numbers are not equal. Call two numbers that are different $x$ and $y$, then $\frac {x^2+y^2} 2$ is more than $(\frac {x+y} 2 )^2$, which is the average if $x$ and $y$ were each replaced by their average, $(\frac {x+y} 2)$.To show this, first expand $(\frac {x+y} 2 )^2$ to $\frac {x^2+y^2+2xy} 4$. Then, to show that $\frac {x^2+y^2} 2>\frac {x^2+y^2+2xy} 4$ multiply by $4$, you get $2x^2+2y^2>x^2+y^2+2xy$. Subtract $x^2+y^2+2xy$ from both sides and get $x^2+y^2-2xy>0$, which is clearly true if $x \ne y$ because it's equal to $(x-y)^2$. This shows that any such "reduction" will always lower the average of squares.
Now, assume that there is a 7-tuple with a lower average than $(3,..,3)$. You can easily reduce it with my method to a 7-tuple with four numbers equal, and the other three equal as well.(i.e. $x,x,x,x,y,y,y$.) Just apply the reduction method to the first two numbers, the 3rd and 4th,1st and 3rd,2nd and 4th,5th and 6th,4th and 7th,4th and 5th,and 6th and 7th, in that order.(Try it with an actual group of distinct numbers if you don't follow:)
Now this 7-tuple's average of square is lower than our original 7-tuple, and so that average must be lower than 9. However this cannot be for the following reason. The sum of the squares in terms of $x$ is: $4x^2+3((\frac {21-4x} 3)^2)$. This expands to $\frac 7 3 (4x^2-24 x+63)$. For this to be less than 63, $7(4x^2-24 x+63)$=$28x^2-168x+441$ must be less than 189, so $28x^2-168x+252=28(x-3)^2$ must be less than zero. This clearly cannot be for any real x.
Therefore, any solution that all the numbers are not the same cannot have an average square of less than that of $(3,..,3)$, namely 9.
QED
A: Let's answer a more general problem: Given numbers $x_1, x_2, \dots,x_n$ such that 
$$
\sum x_i=\sum_{i=1}^n=S
$$
we want to find values $x_i$ such that
$$
\sum x_i^{\,2}
$$
is minimized.
For reasons that will be clear in a moment, consider the sum $\sum(x_i-(S/n))^2$. We'll have
$$\begin{align}
\frac{1}{n}\sum\left(x_i-\frac{S}{n}\right)^2 &= \frac{1}{n}\sum\left(x_i^{\,2}-2\frac{S}{n}x_i+\left(\frac{S}{n}\right)^2\right)\\
&=\frac{1}{n}\sum x_i^{\,2}-\frac{2}{n}\frac{S}{n}\sum x_i+\frac{1}{n}\sum\left(\frac{S}{n}\right)^2\\
&=\frac{1}{n}\sum x_i^{\,2}-2\frac{S}{n}\frac{S}{n}+\frac{1}{n}n\left(\frac{S}{n}\right)^2\\
&=\frac{1}{n}\sum x_i^{\,2}-\left(\frac{S}{n}\right)^2
\end{align}$$
Consequently, we have
$$
\frac{1}{n}\sum x_i^{\,2}=\frac{1}{n}\sum\left(x_i-\frac{S}{n}\right)^2+\left(\frac{S}{n}\right)^2
$$
In other words, the left-hand side, i.e., the average of the squares, will be minimized when the right-hand side is minimized. That clearly will happen when all of the terms $x_i-(S/n)$ are zero, which can only happen when all the $x_i=S/n$, which is to say when each $x_i$ is equal to the average of all the $x$s.
In the original problem, with $n=7$ and $S=21$, that will happen when each $x_i=3$.
A: we need 7 numbers whose squares's average should be minimum, i.e. we need smallest of numbers to be added.
Simply we can observe
21=3+3+3+3+3+3+3
thus average of squares of these 7 numbers = (9+9+9+9+9+9+9)/7=9
A: Consider the formula $V(X)=E[X^2]-(E[X])^2$ for variance, which is necessarily non-negative.
A: $$
{\cal F}
\equiv
{1 \over 7}\sum_{n = 1}^{7}x_{n}^{2} - \mu\left(\sum_{n = 1}^{7}x_{n} - 21\right)
$$
$$
0 = {\partial{\cal F} \over \partial x_{n}}
=
{2 \over 7}\,x_{n} - \mu
\quad\Longrightarrow\quad
x_{n} = {7 \over 2}\,\mu
\quad\Longrightarrow\quad
\mu = {42 \over 49}
\quad\Longrightarrow\quad
x_{n} = 3
$$
$$
{\cal F}
=
{1 \over 7}\sum_{n = 1}^{7}\left(3 + \epsilon_{n}\right)^{2}
=
9
+
{6 \over 7}\overbrace{\sum_{1}^{n}\epsilon_{n}}^{0}
+
{1 \over 7}\sum_{n = 1}^{7}\epsilon_{n}^{2}\quad
\color{#ff0000}{\large\geq\ 9}
$$
