<s>[INST] 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?
 [/INST] 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
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