Tag: Cauchy-Schwarz

Prove the product of sums of n numbers and their reciprocals is greater than n^2

by RoRi

July 11, 2015

Let $x_1, \ldots, x_n \in \mathbb{R}_{>0}$ and let $y_k = \frac{1}{x_k}$ for $k = 1, \ldots, n$ . Prove

$\left( \sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right) \geq n^2.$

Proof. First,

$y_k = \frac{1}{x_k} \quad \implies \quad \sqrt{y_k} = \frac{1}{\sqrt{x_k}} \quad \implies \quad \sqrt{x_k} \sqrt{y_k} = 1$

for all $k = 1, \ldots, n$ . We then apply the Cauchy-Schwarz inequality to the numbers $\sqrt{x_1}, \ldots, \sqrt{x_n}$ and $\sqrt{y_1}, \ldots, \sqrt{y_n}$ . So, Cauchy-Schwarz gives us,

$\begin{align*} &&\left( \sum_{k=1}^n a_k b_k \right)^2 &\leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)\\ \implies && \left( \sum_{k=1}^n \sqrt{x_k} \sqrt{y_k} \right)^2 &\leq \left( \sum_{k=1}^n (\sqrt{x_k})^2 \right) \left( \sum_{k=1}^n (\sqrt{y_k})^2 \right) \\ \implies && \left( \sum_{k=1}^n 1 \right)^2 &\leq \left( \sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right)\\ \implies && n^2 &\leq \left(\sum_{k=1}^n x_k \right) \left( \sum_{k=1}^n y_k \right). \qquad \blacksquare \end{align*}$

Prove the pth power mean is less than the (2p)th power mean

by RoRi

July 10, 2015

We recall the definition of the $p$ th power mean $M_p$ .

For $x_1, \ldots, x_n \in \mathbb{R}_{>0}$ , and $p \in \mathbb{Z}$ with $p \neq 0$ , we define the $p$ th power-mean $M_p$ as:

$M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p}.$

Now, for $p > 0$ , prove $M_p < M_{2p}$ for $x_1, \ldots, x_n$ not all equal.

Proof. From the Cauchy-Schwarz inequality we know that for real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ , we have

$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

with equality if and only if there is some $y \in \mathbb{R}$ such that $a_k y + b_k = 0$ for all $k$ . Letting $a_k = x_k^p$ and $b_k = 1$ we have

$\left( \sum_{k=1}^n x_k^p \cdot 1 \right)^2 < \left( \sum_{k=1}^n x_k^{2p} \right) \left( \sum_{k=1}^n 1 \right).$

This inequality is strict since if equality held there would exist some $y \in \mathbb{R}$ such that $(x_k^p)y + 1 = 0$ for all $k$ , but this would imply $x_k = \left(-\frac{1}{p} \right)^{1/p}$ for all $k$ , contradicting our assumption that the $x_k$ are not all equal. Since $\sum_{k=1}^n 1 = n$ (see here), this implies

$\begin{align*} &&\sum_{k=1}^n x_k^p &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2} n^{1/2} \\ \implies && \left( \sum_{k=1}^n x_k^p \right)^{1/p} &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2p} \cdot n^{1/2p} &(\text{raising to } 1/p)\\ \implies && \left (\frac{1}{n} \right)^{1/p} \cdot \left( \sum_{k=1}^n x_k^p \right)^{1/p} &< \left( \sum_{k=1}^n x_k^{2p} \right)^{1/2p} \cdot \left( \frac{1}{n} \right)^{1/2p} & (\text{multiplying by } (1/n)^{1/p})\\ \implies && \left( \frac{\sum_{k=1}^n x_k^p}{n} \right)^{1/p} &< \left( \frac{\sum_{k=1}^n x_k^{2p}}{n} \right)^{1/2p} & (\text{combining terms})\\ \implies && M_p &< M_{2p}. \qquad \blacksquare \end{align*}$

Prove an if and only if condition for equality in the Cauchy-Schwarz inequality

by RoRi

July 7, 2015

Recall the Cauchy-Schwarz inequality,

For arbitrary real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ we have

$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

The claim is then that the equality sign holds if and only if there is a real number $x$ such that $a_k x + b = 0$ for each $k =1, \ldots, n$ .

Proof. ( $\Rightarrow$ ) If $a_k = 0$ for all $k$ , then equality clearly holds. Assume then that $a_k \neq 0$ for at least one $k$ .

$\left( \sum_{k=1}^n a_k b_k \right)^2 = \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right).$

Then, considering the equation $\sum_{k=1}^n (a_k x + b_k)^2 \geq 0$ and defining,

$A = \sum_{k=1}^n a_k^2, \qquad B = \sum_{k=1}^n b_k a_k, \qquad C = \sum_{k=1}^n b_k^2.$

We have,

$Ax^2 + 2Bx + C \geq 0 \ \implies \ x = \frac{-2B \pm \sqrt{4B^2 - 4AC}}{2A} = \frac{-B \pm \sqrt{B^2 - AC}}{A}.$

But, since we know $B^2 = AC$ (by assumption), we have $x = -\frac{B}{A}$ which is in $\mathbb{R}$ (since $A \neq 0$ since $a_k \neq 0$ for at least one $k$ and each term $a_k^2$ in nonnegative, so the sum is strictly positive).
( $\Leftarrow$ ) Assume there exists $x \in \mathbb{R}$ such that $a_k x + b_k =0$ for each $k =1, \ldots, n$ . Then, $a_k x + b_k = 0 \implies b_k = (-x)a_k$ . So,

$\begin{align*} \left( \sum_{k=1}^n a_k b_k \right)^2 &= \left( -x \left(\sum_{k=1}^n a_k^2 \right) \right)^2 \\ &= x^2 \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n a_k^2 \right) \\ &= \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n x^2 a_k^2 \right) \\ &= \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n (-xa_k)^2 \right) \\ &= \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right). \qquad \blacksquare \end{align*}$