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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 !

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    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. – Hugo Mar 01 '21 at 08:26
  • Copy that, sir. –  Mar 01 '21 at 08:27
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    How is this related to Karamata's inequality? – Martin R Mar 01 '21 at 08:31
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    Btw, you can find the same solution on AoPS: https://artofproblemsolving.com/community/c6h1245073p6388619, and an alternative solution here: https://math.stackexchange.com/q/1792511. – Martin R Mar 01 '21 at 08:34
  • @MartinR, my inspiration_ https://math.stackexchange.com/a/3366590/822157, we will use $a_{1}+ a_{2}, a_{1}+ a_{2}+ a_{3},$ that's much related to $a_{1}\left ( a_{1}+ a_{2}+ a_{3} \right )+ a_{3}\left ( 300- a_{1}- a_{2}- a_{3} \right ),$ I think that. –  Mar 01 '21 at 08:35
  • Roger that, sir –  Mar 01 '21 at 08:35
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    Unless I am mistaken, your estimates work only if all $a_j$ are non-negative. – Martin R Mar 01 '21 at 09:27
  • But $a_{1}, a_{2}, a_{3}> 0$ true. –  Mar 01 '21 at 11:01

2 Answers2

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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.

Hugo
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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 $$ a_1 + \ldots + a_k \ge \frac{a_1 + \ldots + a_n}{k} \, . $$

Martin R
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