With $n> 3,$ prove that $a_{1}+ a_{2}+ a_{3}\geq 100$ by using Karamata's inequality 
Given $n$ real numbers $a_{1}, a_{2}\cdots a_{n}$ so that
$$a_{1}\geq a_{2}\geq\cdots\geq a_{n}, a_{1}+ a_{2}+ \cdots+ a_{n}= 300, a_{1}^{2}+ a_{2}^{2}+ \cdots+ a_{n}^{2}> 10000$$
With $n> 3,$ prove that $a_{1}+ a_{2}+ a_{3}\geq 100.$

I wanna create a contractdiction. So we may assume that
$$a_{1}+ a_{2}+ a_{3}< 100\Rightarrow\sum_{k= 1}^{n}a_{k}^{2}\leq a_{1}\left ( a_{1}+ a_{2}+ a_{3} \right )+ a_{3}\sum_{k= 4}^{n}a_{k}= a_{1}\left ( a_{1}+ a_{2}+ a_{3} \right )+$$
$$+ a_{3}\left ( 300- a_{1}- a_{2}- a_{3} \right )\leq 100\left ( a_{1}- a_{3} \right )+ 300a_{3}\leq 100\left ( a_{1}+ a_{2}+ a_{3} \right )= 10000$$
I tried to solve this inequality by using Karamata but unsuccessfully, who can help me with a way ?? Thanks a real lot !
 A: You are very close to the solution, just change the last equality sign to a $<$ sign because of your assumption $a_1+a_2+a_3<100$, and this contradicts the second part of the hypothesis.
A: You asked for a proof using Karamata's inequality, so here it is. I'll assume that all $a_j$ are non-negative.
Case 1: $3 \le n \le 9$. Then
$$
 300 = a_1 + \cdots + a_n + \underbrace{0 + \cdots + 0}_{9-n \text{ terms}}
\le 3 (a_1+a_2+a_3)
$$
which implies $a_1+a_2+a_3 \ge 100$.
Case 2: $n > 9$. Assume that  $a_1+a_2+a_3 < 100$. Define
$$
b_1, \ldots, b_n = \underbrace{\frac{100}{3}, \ldots, \frac{100}{3}}_{9 \text{ terms}}, \underbrace{0, \ldots, 0}_{n-9 \text{ terms}} \, .
$$
Then
$$
 100 > a_1+a_2+a_3 \ge 3a_1 \implies b_1 \ge a_1\\
 100 > a_1+a_2+a_3 \ge \frac 32 (a_1+a_2) \implies b_1+b_2 \ge a_1+a_2 \\
 100 > a_1+a_2+a_3 \implies b_1+b_2+b_3 \ge a_1+a_2+a_3 \\
\vdots \\
300 = b_1 + \cdots + b_n = a_1 + \cdots + a_n 
$$
which shows that $(b_1, \ldots, b_n)$ majorizes $(a_1, \ldots, a_n)$. Therefore one can apply Karamata's inequality with the convex function $f(x) = x^2$. It follows that
$$
 10\,000 = b_1^2+ \cdots + b_n^2 \ge a_1^2+ \cdots + a_n^2
$$
in contradiction to the assumption.

In the same way one can prove the following generalization:

Let $1 \le k \le n$ be integers and $a_1 \ge a_2 \ge \cdots \ge a_n$ be non-negative real numbers. If
$$
 (a_1 + \ldots + a_n)^2 < k^2 (a_1^2 + \ldots + a_n^2)
$$
then
$$
 a_1 + \ldots + a_k \ge \frac{a_1 + \ldots + a_n}{k} \, .
$$

