# Prove that $a_1 > (a_1 + a_2 + a_3 + a_4)/4$ implies $a_i < (a_1 + a_2 + a_3 + a_4)/4$ for $i \in \{2, 3, 4\}$. [closed]

Given real numbers $a_1, a_2, a_3, a_4$, if $a_1 > (a_1 + a_2 + a_3 + a_4)/4$ , then there exists $i \in \{2, 3, 4\}$ such that $a_i < (a_1 + a_2 + a_3 + a_4)/4$.

I could solve the problem in my mind, but I can't put the solution on the paper. I just have difficulty in writing the solution of the question.

## closed as off-topic by Namaste, José Carlos Santos, Ken Duna, Brian Borchers, Math LoverNov 21 '17 at 16:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, José Carlos Santos, Ken Duna, Brian Borchers, Math Lover
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• Are you asking for a proof of this? All I see is a sentence. Also, you should move the actual question to the body of the post. And if you can "solve the problem in [your] mind", you should at least try to explain how you did so in the post. – Kevin Long Nov 20 '17 at 17:15
• Just to be clear, is it $a_1 + a_2 + a_3 + a_4/4$ or $\frac{a_1 + a_2 + a_3 + a_4}{4}$? – Math Lover Nov 20 '17 at 17:25
• Are you sure you're not missing parentheses around some parts? – user499203 Nov 20 '17 at 17:25
• Yes, I'm asking for a proof. I forgot to use parentheses where it should be like (a1+a2+a3+a4)/4 at both of the inequality. Sorry for asking lack of information. – Hasan Kayhan Nov 20 '17 at 17:30
• suppose it is not true. Quickly you will run into a contradiction. – Doug M Nov 20 '17 at 17:31

Hint: Suppose not. Then we have $$a_2 \ge \frac{a_1+a_2+a_3+a_4}4$$ and the same for $a_3$ and $a_4$. Now add up $a_1+a_2+a_3+a_4$ and see what happens.