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	Suppose you have a list of positive real numbers $a_i$ so that $\sum_{i=1}^n{a_i} = n$. Let $b \geq 0$. Then, $$ \sum_{i=1}^n{a_i^{-b}} \geq \sum_{i=1}^n{a_i^{1-b}} $$ Case 1: $b \leq 1$ Lemma 1: $\sum_{i=1}^n{a_i^{-b}} \geq n$ Let $X$ be a random variable that uniformly selects an $a_i$. Since $x^{-b}$ is concave, we know by Jensen's inequality that $$ E\left[X^{-b}\right] \geq E\left[X\right]^{-b} $$ Furthermore, we know $$ E\left[X^{-b}\right] = \frac{1}{n} \sum_{i=1}^n{a_i^{-b}} $$ and $$ E\left[X\right]^{-b} = \left( \frac{1}{n} \sum_{i=1}^n{a_i} \right)^{-b} = 1^{-b} = 1 $$ Therefore, $$ \frac{1}{n} \sum_{i=1}^n{a_i^{-b}} \geq 1 $$ So, $$ \sum_{i=1}^n{a_i^{-b}} \geq n $$ Lemma 2: $ \sum_{i=1}^n{a_i^{1-b}} \leq n $ The proof of this is similar as that for Lemma 1. However, since $x^{1-b}$ is convex, the inequality flips. By Lemma 1, Lemma 2, and the transitivity of inequality, we achieve and prove $$ \sum_{i=1}^n{a_i^{-b}} \geq \sum_{i=1}^n{a_i^{1-b}} $$ Case 2: $b \gt 1$ In this case, we can rewrite the theorem as $$ \sum_{i=1}^n{a_i^{-c-1}} \geq \sum_{i=1}^n{a_i^{-c}} $$ where $c \gt 0$. Let $$ u = \sum_{i=1}^n{a_i^{-c-1}} - \sum_{i=1}^n{a_i^{-c}} $$ then $$ u = \sum_{i=1}^n{a_i^{-c-1} - a_i^{-c}}$$ Consider the derivative: $$ \frac{du}{dc} = \sum_{i=1}^n{\ln{a_i} \cdot a_i^{-c} - \ln{a_i} \cdot a_i^{-c-1}}$$ so $$ \frac{du}{dc} = \sum_{i=1}^n{\ln{a_i} \cdot \left( a_i^{-c} - \cdot a_i^{-c-1} \right)}$$ Lemma 3: Each term in this derivative sum is positive. Case a: $a_i \lt 1$ In this case, both $\ln{a_i}$ and $ a_i^{-c} - \cdot a_i^{-c-1} $ are negative, so the overall term is positive. Case b: $a_i = 1$ In this case, both $\ln{a_i}$ and $ a_i^{-c} - \cdot a_i^{-c-1} $ are zero, so the overall term is zero. Case c: $a_i \gt 1$ In this case, both $\ln{a_i}$ and $ a_i^{-c} - \cdot a_i^{-c-1} $ are positive, so the overall term is positive. Since we know $u$ is non-negative for $c=0$ (i.e. $b=1$) and we know the derivative is always non-negative, we can conclude that $u$ is always non-negative. By case analysis, we've proven the theorem.

Suppose you have a list of positive real numbers $a_i$ so that $\sum_{i=1}^n{a_i} = n$. Let $b \geq 0$. Then, $$ \sum_{i=1}^n{a_i^{-b}} \geq \sum_{i=1}^n{a_i^{1-b}} $$

Case 1: $b \leq 1$

Lemma 1: $\sum_{i=1}^n{a_i^{-b}} \geq n$

Let $X$ be a random variable that uniformly selects an $a_i$. Since $x^{-b}$ is concave, we know by Jensen's inequality that $$ E\left[X^{-b}\right] \geq E\left[X\right]^{-b} $$ Furthermore, we know $$ E\left[X^{-b}\right] = \frac{1}{n} \sum_{i=1}^n{a_i^{-b}} $$ and $$ E\left[X\right]^{-b} = \left( \frac{1}{n} \sum_{i=1}^n{a_i} \right)^{-b} = 1^{-b} = 1 $$ Therefore, $$ \frac{1}{n} \sum_{i=1}^n{a_i^{-b}} \geq 1 $$ So, $$ \sum_{i=1}^n{a_i^{-b}} \geq n $$

Lemma 2: $ \sum_{i=1}^n{a_i^{1-b}} \leq n $

The proof of this is similar as that for Lemma 1. However, since $x^{1-b}$ is convex, the inequality flips.

By Lemma 1, Lemma 2, and the transitivity of inequality, we achieve and prove $$ \sum_{i=1}^n{a_i^{-b}} \geq \sum_{i=1}^n{a_i^{1-b}} $$

Case 2: $b \gt 1$

In this case, we can rewrite the theorem as $$ \sum_{i=1}^n{a_i^{-c-1}} \geq \sum_{i=1}^n{a_i^{-c}} $$ where $c \gt 0$.
Let $$ u = \sum_{i=1}^n{a_i^{-c-1}} - \sum_{i=1}^n{a_i^{-c}} $$ then $$ u = \sum_{i=1}^n{a_i^{-c-1} - a_i^{-c}}$$ Consider the derivative: $$ \frac{du}{dc} = \sum_{i=1}^n{\ln{a_i} \cdot a_i^{-c} - \ln{a_i} \cdot a_i^{-c-1}}$$ so $$ \frac{du}{dc} = \sum_{i=1}^n{\ln{a_i} \cdot \left( a_i^{-c} - \cdot a_i^{-c-1} \right)}$$

Lemma 3: Each term in this derivative sum is positive.

Case a: $a_i \lt 1$

In this case, both $\ln{a_i}$ and $ a_i^{-c} - \cdot a_i^{-c-1} $ are negative, so the overall term is positive.

Case b: $a_i = 1$

In this case, both $\ln{a_i}$ and $ a_i^{-c} - \cdot a_i^{-c-1} $ are zero, so the overall term is zero.

Case c: $a_i \gt 1$

In this case, both $\ln{a_i}$ and $ a_i^{-c} - \cdot a_i^{-c-1} $ are positive, so the overall term is positive.

Since we know $u$ is non-negative for $c=0$ (i.e. $b=1$) and we know the derivative is always non-negative, we can conclude that $u$ is always non-negative.

By case analysis, we've proven the theorem.