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### Some Applications of CSB Inequality
The first application of the CSB inequality is the _triangle inequality_, which is also extremely useful.
**Theorem 69** (The Triangle Inequality).: _Let \(u\) and \(v\) be vectors in an inner product space \(V\). Then_
\[||u+v||\leq||u||+||v||.\]
Proof
: Let \(u,v\in V\).
\[||u+v||^{2} = \langle u+v,u+v\rangle\] \[= \langle u,u\rangle+\langle v,v\rangle+\langle u,v\rangle+\langle v,u\rangle\] \[= ||u||^{2}+||v||^{2}+\langle u,v\rangle+\overline{\langle u,v\rangle}\] \[= ||u||^{2}+||v||^{2}+2\mathrm{Re}\langle u,v\rangle\] \[\leq ||u||^{2}+||v||^{2}+2|\langle u,v\rangle|\] \[\leq ||u||^{2}+||v||^{2}+2||u||\cdot||v||=(||u||+||v||)^{2},\]
where we have used the CSB inequality
\[2|\langle u,v\rangle|\leq 2||u||\cdot||v||.\]
This implies that
\[||u+v||\leq||u||+||v||.\qed\]If we consider \(x,y\in\mathbb{R}^{n}\) with the Euclidean norm, the CSB inequality takes on the form:
\[\sum_{i=1}^{n}|x_{i}y_{i}|\leq\left(\sum_{i=1}^{n}|x_{i}|^{2}\right)^{1/2}\left( \sum_{i=1}^{n}|y_{i}|^{2}\right)^{1/2},\]
or in our notation we have:
**Theorem 70** (Cauchy’s Inequality).: _Let \(x,y\in\mathbb{R}^{n}\) with the Euclidean norm. Then_
\[\sum_{i=1}^{n}|x_{i}y_{i}|\leq||x||\cdot||y||.\]
Note that we may suppose that \(x\neq 0\), \(y\neq 0\) in the above Theorem 70; there is nothing to show if either is \(0\).
**Remark 38** (Schwarz Inequality).: The integral analog of Cauchy’s inequality is known as the Schwarz inequality. To explain Schwarz Inequality we need to define the set \(L^{2}[a,b]\) and understand the inner product on this space. \(L^{2}[a,b]\) denote the space of square Lebesgue integrable functions, Lebesgue integrable functions will be defined later in the text, see Definition 98.
At this point we just want to observe that for \(f,g\in L^{2}[a,b]\) if the inner product given as below, then we have
\[|\langle f,g\rangle| = \left|\int_{a}^{b}f(x)g(x)dx\right|\leq\int_{a}^{b}|f(x)|\cdot|g (x)|dx\] \[\leq \left(\int_{a}^{b}|f(x)|^{2}\right)^{1/2}\left(\int_{a}^{b}|g(x) |^{2}\right)^{1/2}.\]
Using the CSB inequality, we can prove interesting inequalities.
Let \(x\) and \(y\) be non-negative real numbers. Applying the CSB inequality to
\[u=\left[\begin{array}{c}\sqrt{x}\\ \sqrt{y}\end{array}\right]\quad v=\left[\begin{array}{c}\sqrt{y}\\ \sqrt{x}\end{array}\right]\]
yields a useful inequality:
**Theorem 71** (AM-GM Inequality).: _Let \(x,y\in\mathbb{R}_{+}\) (positive real numbers). Then,_
\[\sqrt{xy}\leq\frac{x+y}{2}.\]
Proof
: We know that
\[|\langle u,v\rangle| = 2\sqrt{xy},\] \[\left|\left|u\right|\right|^{2} = x+y,\] \[\left|\left|v\right|\right|^{2} = x+y,\]and \(x\) and \(y\) are non-negative implying that \(||u||=||v||=\sqrt{x+y}\). Therefore,
\[2\sqrt{xy}=||\langle u,v\rangle||\leq||u||\cdot||v||=\sqrt{x+y}\sqrt{x+y}=x+y.\]
The right side of AM-GM inequality is the familiar arithmetic mean of \(x,y\) and the left side is the geometric mean.
**Corollary 13**.: _The AM-GM inequality holds for \(n\) non-negative variables, \(x_{1},\)\(x_{2},\ldots,x_{n}\). It has the form:_
\[\sqrt[n]{x_{1}x_{2}\cdots x_{n}}\leq\frac{x_{1}+x_{2}+\cdots+x_{n}}{n},\]
_with equality if and only if \(x_{1}=x_{2}=\cdots=x_{n}\)._
**Lemma 8** (Minkowski’s Inequality).: _Let \(x,y\in\mathbb{R}^{n}\) or \(\mathbb{C}^{n}\). Then_
\[||x+y||\leq||x||+||y||,\]
_or equivalently_
\[\left(\sum_{i=1}^{n}|x_{i}+y_{i}|^{2}\right)^{1/2}\leq\left(\sum_{i=1}^{n}|x_{ i}|^{2}\right)^{1/2}+\left(\sum_{i=1}^{n}|y_{i}|^{2}\right)^{1/2}.\]
Proof
: Using Cauchy’s inequality,
\[||x+y||^{2} = \sum_{i=1}^{n}|x_{i}+y_{i}|\cdot|x_{i}+y_{i}|\leq\sum_{i=1}^{n}|x _{i}+y_{i}|(|x_{i}|+|y_{i}|)\] \[= \sum_{i=1}^{n}|x_{i}+y_{i}|\cdot|x_{i}|+\sum_{i=1}^{n}|x_{i}+y_{i }|\cdot|y_{i}|\] \[\leq ||x+y||\cdot||x||+||x+y||\cdot||y||=||x+y||(||x||+||y||).\]
This implies that
\[||x+y||^{2}\leq||x+y||(||x||+||y||).\]
If \(||x+y||=0\), then the above lemma is trivially true. Otherwise, Minkowski’s inequality follows from the inequality last written by dividing through by \(||x+y||\).
**Remark 39**.: Observe that Lemma 8 is a special case of the triangle inequality given in Theorem 69. It illustrates the triangle inequality for \(x,y\in\mathbb{R}^{n}\) or \(\mathbb{C}^{n}\).
Suppose a given inner product space has been normed by taking
\[||w||=\sqrt{\langle w,w\rangle}.\]
We say a sequence \(\{x_{n}\}\) converges to \(x\) when \(||x_{n}-x||\to 0\). Since every inner product gives rise to a norm, we can talk about convergent sequences or Cauchy sequences in an inner product space as well. The following is such an example:
**Theorem 72**.: _If \(\{x_{n}\}\) and \(\{y_{n}\}\) are sequences in an inner product space, which converge to \(x\) and \(y\), respectively, then \(\{\langle x_{n},y_{n}\rangle\}\) is convergent sequence with limit \(\langle x,y\rangle\)._
Proof
: To prove \(\langle x_{n},y_{n}\rangle\rightarrow\langle x,y\rangle\), we write
\[\langle x_{n},y_{n}\rangle-\langle x,y\rangle = \langle x_{n},y_{n}\rangle-\langle x_{n},y\rangle+\langle x_{n}, y\rangle-\langle x,y\rangle\] \[= \langle x_{n},y_{n}-y\rangle+\langle x_{n}-x,y\rangle,\]
so that by using the CSB inequality we obtain:
\[|\langle x_{n},y_{n}\rangle-\langle x,y\rangle| \leq |\langle x_{n},y_{n}-y\rangle|-|\langle x_{n}-x,y\rangle|\] \[\leq ||x_{n}||||y_{n}-y||+||x_{n}-x||||y||.\]
Every convergent sequence is bounded, so \(||x_{n}||\leq K\) for some constant \(K\) and for all \(n\). Moreover \(||x_{n}-x||\to 0\) and \(||y_{n}-y||\to 0\), hence, as claimed, \(\langle x_{n},y_{n}\rangle\rightarrow\langle x,y\rangle\).
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