Tricky Multivariable Inequality Given 100 positive real numbers $x_1, x_2, \cdots, x_n$ that satisfy
$x_1^2+x_2^2+\cdots+x_n^2>10000$ and
$x_1+x_2+\cdots x_n\le 300$,
prove that there exist three numbers from this set such that the sum of these three numbers is larger than 100.
I have tried using Lagrange Multipliers and smoothing but it appears not to work. An "equality case" is $(100/3,...$(9 times)$..., 100/3,0,0,0...)$
 A: I shall prove the following statement, from which the problem is solved by setting $B=300$ and $n=100$.
Let $S=\{x_1,x_2,\ldots,x_n\}$ be a set of $n \geq 3$ non-negative numbers and let $t=\sum_{i=1}^{n}x_i^2$ and 
$s=\sum_{i=1}^{n} x_i$.
Suppose that $s \geq \dfrac{B^2}{9}$ and $t \leq B$. Then there exist three numbers in $S$ whose sum is at least 
$\dfrac{B}{3}$.
Proof:
Let $A=\dfrac{B^2}{9}$, $p_i=\dfrac{x_i}{s}$.
We have: $\sum_{i=1}^{n}{p_i}=1$ and $\sum p_ix_i= \dfrac{t}{s} \geq \dfrac{A}{B}$.
Hence we can find a number $x$ in the set of value at least $\dfrac{A}{B}$.
Repeating this argument, we can find two more numbers $y,z$ such that:
$y \geq \dfrac{A-x^2}{B-x}$ and $z \geq \dfrac{A-x^2-y^2}{B-x-y}$.
We need to prove:
 $x+y+z \geq 3\dfrac{A}{B}$,
for which I'll use the following claims.
Claim 1: Let $A,B> 0$ such that $B^2>A$ and let $f(x)=x+\dfrac{A-x^2}{B-x}$ and $g(x)=x+2\dfrac{A-x^2}{B-x}$.
Then $f$ is increasing when $x \leq B-\sqrt{\dfrac{B^2-A}{2}}$ and $g$ is increasing when $x \leq 2B-\sqrt{3B^2-2A}$.
In particular, if $B^2=9A$, then $f,g$ are both increasing when $x \leq \dfrac{B}{3}$.
This claim is proved by differentiating $f,g$ and some algebra.
Claim 2: Let $A,B >0$, $A \leq B^2 \leq \dfrac{4A}{3}$ and $\sqrt{A} \geq x \geq \dfrac{A}{B}$; 
then $f(x)=x+\dfrac{A-x^2}{B-x} \geq 2\dfrac{A}{B}$, with the minimum at $x=\dfrac{A}{B}$.
Proof of Claim 2: If $x \geq 2\dfrac{A}{B}$, we are done because the second term is non-negative,
so let $x \leq 2\dfrac{A}{B}$. Using claim 1 for $f$ and some algebra, we see that $f$ is increasing in this range of $x$;
hence the minimum is at $x=\dfrac{A}{B}$.
Now we prove the original desired statement that $x+y+z \geq 3\dfrac{A}{B}$.
If $x \geq \dfrac{B}{3}=3\dfrac{A}{B}$, we are done, so let's assume that $x \leq \dfrac{B}{3}$.
Consider $y+ \dfrac{A-x^2-y^2}{B-x-y}$: this is at least $2\dfrac{A-x^2}{B-x}$ by applying Claim 2 and 
checking that its preconditions are met.
Thus $x+y+z \geq x+2\dfrac{A-x^2}{B-x}$.
Again using Claim 1 for $g$, we see that $g$ is increasing in the range of $x$ and hence its minimum is at $x=\dfrac{A}{B}$.
This gives $x+y+z \geq 3\dfrac{A}{B}=\dfrac{B}{3}$, which completes the proof.
