# Cauchy Schwarz inequality

Let $a_1,a_2,\ldots ,a_n$, be positive real numbers and S = $a_1+a_2+a_3+\cdots+a_n$. Use the Cauchy-Schwarz inequality to prove that

$$\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}$$ and $$\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)$$

Hint: Apply the inequality to $\sum \frac{S}{a_i}$ and $\sum \frac{a_i}{S}$ for one case and $\sum a_i(S-a_i)$ and $\sum \frac{a_i}{S-a_i}$ for the other case.

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Have you written down the version of Cauchy-Schwarz you want to use? If not, do it! – martini Mar 27 '12 at 10:08
I think the hint says clearly what you have to do. You only need to write the corresponding Cauchy Schwarz inequality and you are done. – Beni Bogosel Mar 27 '12 at 10:17

## 2 Answers

Cauchy Schwarz Inequality:

\begin{align*} \left( \frac{1}{a_1}+\frac{1}{a_2}+\dots +\frac{1}{a_n}\right)\left( a_1+a_2+\dots+a_n \right) &\geq \left(\frac{1}{\sqrt{a_1}}\sqrt{a_1}+\frac{1}{\sqrt{a_2}}\sqrt{a_2}+\dots +\frac{1}{\sqrt{a_n}}\sqrt{a_n} \right)^2 \\\ &\geq \left( 1+1+\dots+1\right)^2 = n^2 \end{align*}

which when applied here \begin{align*} \displaystyle\sum_{i=1}^n \frac{S}{a_i} &= \left( \frac{1}{a_1}+\frac{1}{a_2}+\dots +\frac{1}{a_n}\right)S \geq n^2\\ \end{align*}

Therefore \begin{align*} \displaystyle\sum_{i=1}^n \frac{S-a_i}{a_i} &= \displaystyle\sum_{i=1}^n \frac{S}{a_i} -n\\ &\geq n^2 - n = n(n-1) \tag{1} \end{align*}

Similarly, re-writing the other summation

\begin{align*} \displaystyle\sum_{i=1}^n \frac{a_i}{S-a_i} &= \frac{S-\displaystyle\sum_{j \neq 1}a_j}{\displaystyle\sum_{j \neq 1}a_j} + \frac{S-\displaystyle\sum_{j \neq 2}a_j}{\displaystyle\sum_{j \neq 2}a_j}+\dots +\frac{S-\displaystyle\sum_{j \neq n}a_j}{\displaystyle\sum_{j \neq n}a_j}\\ &= \displaystyle\sum_{i=1}^n \left(\frac{S}{\displaystyle\sum_{j \neq i}a_j}\right)-n \tag{2} \end{align*}

Also since

$$\displaystyle\sum_{j \neq 1}a_j+\displaystyle\sum_{j \neq 2}a_j+\dots+\displaystyle\sum_{j \neq n}a_j = (n-1)S$$

By Cauchy-Schwarz Inequality \begin{align*} \displaystyle\sum_{i=1}^n \left(\frac{1}{\displaystyle\sum_{j \neq i}a_j}\right)(n-1)S \geq n^2\\ \Longrightarrow \displaystyle\sum_{i=1}^n \left(\frac{1}{\displaystyle\sum_{j \neq i}a_j}\right) \geq \frac{n^2}{(n-1)S} \end{align*}

From $(2)$

\begin{align*} \displaystyle\sum_{i=1}^n \frac{a_i}{S-a_i} &\geq \left(\frac{n^2}{n-1}\right) - n\\ &\geq \frac{n}{n-1} \end{align*}

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The left sides are the values of $\sum f(p_i)$ and $\sum g(p_i)$ with $\sum p_i = 1$ being constant, where

$f(x) = \dfrac{x} {1-x}$ $g(x) = \dfrac 1 x - 1$, and $p_i = \dfrac {a_i} S$.

Both functions are convex for $x>0$, so that each sum of values is minimized when all arguments $p_i$ (and therefore all $a_i$) are equal.

Another line of argument that makes this obvious is the Rearrangement Inequality: the average of the sequence $x_i y_i$ is larger or equal to the product of the averages of $x_i$ and $y_i$ if the two sequences are in the same order. In the first inequality the sequences $a_i$ and $(S-a_i)^{-1}$ are in the same order, and in the second inequality, $(S-a_i)$ and $1/a_i$ are in the same order. The averages of $a_i$ and $(S-a_i)$ are $S/n$ and $(n-1)S/n$, which depend only on $S$. The other averages lead to the standard problem of minimizing a sum of reciprocals given the sum of the variables, which happens when the summands are equal.

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@Peter , I asked a meta question about \Sigma vs \sum . – zyx Apr 9 '12 at 21:17
Link? What is the issue? – Pedro Tamaroff Apr 9 '12 at 21:18
@Peter, meta.math.stackexchange.com/questions/3932/… . No issue, a question. – zyx Apr 9 '12 at 21:21