Prove $\exists x_m + x_n + x_p \ge 100$ 
Let $x_i \in \mathbb {R^+}$ (with $i = \overline{1,2,3,\ldots,100}$)
  such that: $$x_1^2+x_2^2+x_3^2+\ldots+x_{100}^2 > 10^4 \land x_1+x_2+x_3 + \ldots+x_{100}<300$$
Prove that there exists $x_m+x_n+x_p \ge 100$ 
(with $m,n,q \in \{1,2,3,\ldots,100\} \land (m-n)(m-p)(n-p) \ne 0\ $)

I see this problem on a math forum. It is nice but I don't know where to start, I can't solve it.
 A: Wlog. $x_1\le x_2\le \ldots \le x_{100}$. Assume $x_{98}+x_{99}+x_{100}< 100$. 
Note that replacing $x_i$ with $x_i-e$ and $x_j$ with $x_j+e$ where $0\le e\le x_i\le x_j$ leaves the linear sum unchanged while increasing the square sum:
$$(x_i-e)^2+(x_j+e)^2 = (x_i^2+x_j^2)+2e^2+2e(x_j-x_i) \ge x_i^2+x_j^2.$$
Likewise we can replace $x_{98},x_{99},x_{100}$ by their mean without destroying the conditions, i.e. we assume $x_{98}=x_{99}=x_{100}=:a$ with $a<\frac{100}3$.
Now run the following algotrithm: 


*

*Let $i$ be minimal with $x_i>0$. Let $j$ be maximal with $x_j<a$.

*If $i\ge j$, terminate.

*Otherwise, let $e=\min\{a-x_j,x_i\}$ and replace $x_i$ by $x_i-e$ and $x_j$ by $x_j+e$. As observed above, we obtain a new ordered sequence of nonnegative numbers with $\sum x_i^2>10^4$, $\sum x_i<300$ and $x_{98}+x_{99}+x_{100}=3a< 100$.

*Go back to step 1.


Upon termination, we have 
 $$x_1=\ldots=x_{i-1}=0<x_i\le x_{i+1}=\ldots=x_{100}=a,$$
hence (with $b:=x_i$ and $n:=100-i$)
$$\tag1 b+na<300$$
$$\tag2 b^2+na^2>10000$$
$$\tag3 0\le b\le a<\frac{100}3$$
From this we get $10000-ab>(300-b)\cdot a = na^2>10000-b$, hence $b=0$ or $a<1$. But if $0\le b\le a<1$, we find $10000<b^2+na^2<b+na<300$, contradiction.
Hence we conclude $b=0$.
Then $n\cdot 10000<(na)^2<90000$ implies $n<9$.
Finally, we get $(3a)^2>na^2>10000$, i.e. $3a>100$, contradiction.
A: Suppose $x_1\ge x_2 \ge x_3\ge \dots \ge x_{100}$, and $x_1+x_2+x_3=s < 100$. Then 
$$
x_1^2+x_2^2+x_3^2+\dots +x_{100}^2\le x_1^2+x_2^2+x_3^2+\lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2 
$$
since it attains the maximal at $(x_1, x_2, x_3, \dots, x_3,x_k,0,\dots)$, and
$$
x_1^2+x_2^2+x_3^2+\lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2 \\
\le (s-2x_3)^2+2x_3^2+ \lfloor\frac{300-s}{x_3}\rfloor x_3^2+(300-\lfloor\frac{300-s}{x_3}\rfloor x_3)^2,
$$
so the problem becomes 
If $a\ge b\ge c \ge 0, n\le 98$, $a+2b<100$ and $a+nb+c<300$, then $a^2+n b^2+c^2<100^2$.
