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.

#Algebra #Power #Inequality #PowerMeanInequality

Easy Math Editor

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Comments

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 - 6 years, 7 months ago

Good manipulation and follow-through.

Note that the condition of $x, y$ are integers is not needed.

Calvin Lin Staff - 6 years, 7 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?

Calvin Lin Staff - 6 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$