Minimum value of algebraic expression. 
If $0\leq x_{i}\leq 1\;\forall i\in \left\{1,2,3,4,5,6,7,8,9,10\right\},$ and $\displaystyle \sum^{10}_{i=1} x^2_{i}=9$
Then $\max$ and $\min$ value of $\displaystyle \sum^{10}_{i=1} x_{i}$

$\bf{My\; Try::}$
Using Cauchy-Schwarz Inequality
$$\left(x^2_{1}+x^2_{2}+.......+x^2_{10}\right)\cdot (1^2+1^2+....1^2)\geq \left(x_{1}+x_{2}+....+x_{10}\right)^2$$
So we get $$\left(x_{1}+x_{2}+....+x_{10}\right)\leq \sqrt{90}$$
Now How can I calculate for its minimum value, Help required, Thanks
 A: Hint: If $0\leq x,y\leq 1$, then $$x+y\geq \sqrt{x^2+y^2}\,,$$ whose equality case is $x=0$ or $y=0$.  If $x^2+y^2\geq1$, we also have $$x+y\geq1+\sqrt{x^2+y^2-1}\,,$$ whose equality case is $x=1$ or $y=1$.
A: Clearly the minimum values is $9$ (since $a_i\geq a_i^2$), we can obtain it by taking $9$ values of $a_i$ to be $1$ and the rest to be $0$.
We flip the question around:
Given non negative reals $a_1,a_2\dots a_n$ with $a_1+a_2+\dots + a_n=A$
What is the minimum possible value of $a_1^2+a_2^2+\dots + a_n^2$?
Since the function $f(x)=x^2$ is convex the minimum possible value is reached when all values are equal, so the minimum is $\frac{A^2}{n}$
So if $a_1+a_2+\dots + a_n> 3\sqrt{n}$ we have $a_1^2+\dots + a_n^2>\frac{9n}{n}=9$
The maximum value is therefore $3\sqrt{n}$, reachable when each number is $\frac{3}{\sqrt{n
}}$
A: You have already got the right inequality for the maximum, all you need to add is that equality can be achieved when $x_i = 3/\sqrt{10}$. 
For the minimum, note $x_i\in [0,1]\implies x_i^2\leqslant x_i\implies 9=\sum x_i^2\leqslant \sum x_i$  Equality is possible here when one of the $x_i$ is $0$ and all others $1$. 
