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.

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 $a^2+n b^2+c^2<100^2$.
