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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, 2 months ago

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

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 $$i$$. · 2 years, 2 months ago

Good manipulation and follow-through.

Note that the condition of $$x, y$$ are integers is not needed. Staff · 2 years, 2 months ago

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? Staff · 2 years, 2 months ago