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Not Quite Power Mean

Given that \(a_1a_2\cdots a_n=1\) and integers \(x,y\) such that \(x\le y\)

Prove that \[a_1^x+a_2^x+\cdots +a_n^x\le a_1^y+a_2^y+\cdots +a_n^y\]

EXTENSION: Generalize the condition to \(\displaystyle\sum_{sym}a_1a_2\cdots a_k=1\) where \(k\) is any integer from \(1\) to \(n\) inclusive.

Note by Daniel Liu
2 years, 4 months ago

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By Power Mean theorem

\[\large\sqrt[x]{\dfrac{1}{n}\sum a_i^x}\le\sqrt[y]{\dfrac{1}{n}\sum a_i^y}~\Longleftrightarrow ~\sum a_i^y\ge \dfrac{n}{n^{y/x}}\left(\sum a_i^x\right)^{y/x}\]

so it suffices to prove that

\[\large\dfrac{n}{n^{y/x}}\left(\sum a_i^x\right)^{y/x}\ge \sum a_i^x ~\Longleftrightarrow ~\left(\sum a_i^x\right)^{(y-x)/x}\ge n^{(y-x)/x}~\Longleftrightarrow ~\sum a_i^x\ge n\]

trivial by AM-GM. Equality holds when \(a_i=1\) for all \(i\). Jubayer Nirjhor · 2 years, 4 months ago

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@Jubayer Nirjhor Good manipulation and follow-through.

Note that the condition of \(x, y \) are integers is not needed. Calvin Lin Staff · 2 years, 4 months ago

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Because I'm a fan of just using AM-GM (esp for those starting out with inequalities), I'd like to mention that there is a way to use AM-GM "directly" (with slight creativity).

Hint: Multiply the LHS by \( 1 = \left( \prod a_i \right) ^ { \frac{y-x}{n} } \).

Can anyone complete this from here? Calvin Lin Staff · 2 years, 4 months ago

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