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Sir , can you please suggest some name for the problem I mentioned above so that I can get some idea and I can try to give names that are meaningful from next time.

I was reading the wiki On AM - GM Inequality .In it I found out that there was a proof which said that it requires a bit of combinatorics . In the last bit of the proof , it said

$k > m_g > 0 \Rightarrow m_g \in {(0 , k)}$

$k \geq m_a > 0 \Rightarrow m_a \in {(0 , k]}$.

From these how does it follow that $m_a \geq m_g$ ?

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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## Comments

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TopNewestWe have removed that functionality.

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Sir, So will you write the feedback on the solution or is the feature removed?

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The functionality of "request feedback from challenge master" has been removed.

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@Calvin Lin Sir , recently I posted two questions ..the problem name changed automatically.

Like I named the problem 'Strange' , it changed to 'A geometry problem by Ankit Kumar Jain.

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We remove titles that are meaningless to the problem.

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Okay sir .

Sir , can you please suggest some name for the problem I mentioned above so that I can get some idea and I can try to give names that are meaningful from next time.

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@Calvin Lin Please help me out!

I was reading the wiki On AM - GM Inequality .In it I found out that there was a proof which said that it requires a bit of combinatorics . In the last bit of the proof , it said

$k > m_g > 0 \Rightarrow m_g \in {(0 , k)}$

$k \geq m_a > 0 \Rightarrow m_a \in {(0 , k]}$.

From these how does it follow that $m_a \geq m_g$ ?

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@Pi Han Goh Can you please help me out?

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I don't know what $k, m_a$ and $m_g$ represents. So I can't help you with that...

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$m_a$ is the arithmetic mean and $m_g$ is geometric mean and $k$ is some constant.

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AM-GM .In the proof that says 'It requires a bit of combinatorics' , that one.

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