10.4 Important inequalities

If we know particular inequalities are true, we can use them to prove other inequalities. The following general inequalities are very useful.

10.4.1 The Triangle Inequality

The Triangle Inequality relates the absolute value of a sum of two numbers to the sum of the absolute values (or modulus) of those two numbers. This makes it very useful when calculating and manipulating quantities involving the absolute value.

Theorem 10.18 (Triangle Inequalities).

For any real numbers $a,b\in\mathbb{R}$

1.

$|a+b|\leq|a|+|b|,$ (10.2)
2.

$|a-b|\geq||a|-|b||.$ (10.3)

Proof.

(1) The proof relies only on careful algebra as follows. For $a,b\in\mathbb{R}$ , notice that

ab\leq|ab|,\mbox{ so that }ab\leq|a|\ |b|.

Indeed, if $ab<0$ , we have $|ab|=-ab>0$ and thus, $ab<0<|ab|$ . On the other hand, if $ab\geq 0$ , then $|ab|=ab$ and hence, $ab\leq|ab|$ . Thus

a^{2}+2ab+b^{2}\leq a^{2}+2|a|\ |b|+b^{2}=|a|^{2}+2|a|\ |b|+|b|^{2}.

Factoring this gives

|a+b|^{2}\leq(|a|+|b|)^{2}

Taking the positive square root on each side

|a+b|\leq|a|+|b|.

(2) The second part is proved as follows: for $a,b\in\mathbb{R}$

ab\leq|ab|,\mbox{ so that }ab\leq|a|\ |b|.

Thus

a^{2}-2ab+b^{2}\geq a^{2}-2|a|\ |b|+b^{2}=|a|^{2}-2|a|\ |b|+|b|^{2}.

Factoring this gives

|a-b|^{2}\geq(|a|-|b|)^{2}

Taking the positive square root on each side

|a-b|\geq||a|-|b||.

Note that if $a,b\geq 0$ , then

|a+b|=a+b=|a|+|b|.

Similarly, if $a,b<0$ , then

|a+b|=-(a+b)=-a-b=|a|+|b|.

Exercise 10.19.

Determine what conditions $a$ and $b$ need to satisfy for (10.3) to be an equality.

Solution (please try for yourself before looking)

We are trying to solve

|a-b|=||a|-|b||.

Note that if $a,b\geq 0$

||a|-|b||=|a-b|.

Similarly if $a,b\leq 0$

||a|-|b||=|-a+b|=|a-b|.

Equality holds if and only if $a$ and $b$ have the same algebraic sign.

This is one of many triangle inequalities that you will see in your degree, so called because they reflect the geometric result that the sum of the lengths of any two sides of a triangle is greater than or equal to the length of the third side.

Figure 10.3: A triangle whose third side is shorter than the sum of the other two sides

Example 10.20.

Suppose we wish to prove that

\left|\frac{5n^{2}+2n}{n^{2}+4}-5\right|\leqslant\frac{22}{n}

for all $n\in\mathbb{N}$ .

By the triangle inequality,

\left|\frac{5n^{2}+2n}{n^{2}+4}-5\right|=\left|\frac{2n-20}{n^{2}+4}\right|% \leq\left|\frac{2n}{n^{2}+4}\right|+\left|\frac{-20}{n^{2}+4}\right|=\frac{2n}% {n^{2}+4}+\frac{20}{n^{2}+4}

for all $n\geqslant 1$ . So since

\frac{2n}{n^{2}+4}+\frac{20}{n^{2}+4}\leq\frac{2n}{n^{2}}+\frac{20}{n}=\frac{2% 2}{n},

we have

\left|\frac{5n^{2}+2n}{n^{2}+4}-5\right|\leq\frac{22}{n}

for all integers $n\geq 1$ .

10.4.2 Bernoulli’s inequality

Bernoulli’s inequality has a lot of applications in real analysis²² 2 You will see it used in IMA. It has a number of variants. Its most general statement is as follows.

Theorem 10.21 (Bernoulli’s Inequality).

Assume $r$ is a real number and $r\geq 1$ . Then

(1+x)^{r}\geq 1+rx\quad\text{for all}\quad x\geq-1.

(10.4)

The proof of this general statement is beyond the scope of this course. However, we are equipped to prove a more restricted version of Bernoulli’s Inequality.

Exercise 10.22.

Prove Bernoulli’s inequality for $r\in\mathbb{N}$ by induction.

Solution (please try for yourself before looking)

Let $P(n)$ be the statement

(1+x)^{n}\geq 1+nx\quad\text{for all}\quad x\geq-1.

For $n=1$ we have

1+x\geq 1+x\quad\text{for all}\quad x\geq-1.

Hence $P(1)$ is true.

Assume $P(k)$ is true and consider $(x+1)^{k+1}$ :

	$\displaystyle(1+x)^{k+1}$	$\displaystyle=(1+x)^{k}(1+x)$
		$\displaystyle\geq(1+kx)(1+x)\text{, since $1+x\geq 0$},$
		$\displaystyle=1+kx+x+kx^{2}$
		$\displaystyle=1+x(k+1)+kx^{2}$
		$\displaystyle\geq 1+(k+1)x.$

Hence $P(k+1)$ is true.

Since $P(1)$ holds and $P(k)\Rightarrow P(k+1)$ , we conclude $P(n)$ holds for all $n\in\mathbb{N}$ by induction.

Notice that $y=1+rx$ is the tangent line to $y=(1+x)^{r}$ at the point $x=0$ . Indeed if we take $f(x)=(x+1)^{r}$ and differentiate $f(x)$ with respect to $x$ , we obtain

f^{\prime}(x)=r(1+x)^{r-1}.

Evaluating this at $x=0$ then gives $f^{\prime}(0)=r$ , which is the gradient of the tangent line at $x=0$ .

Since the tangent line to $y=f(x)$ at $x=x_{0}$ is given by

y=f(x_{0})+f^{\prime}(x_{0})(x-x_{0}),

we get our tangent line

g(x)=1+nx,

by substituting our value $x_{0}=0$ .

For a specific value of $r$ , we can find a larger range of $x$ for which Bernoulli’s inequality holds by calculating the intersection point between this tangent line and our graph of $y=(1+x)^{r}$ . For example, when $r=3$ , we can bring the lower bound for $x$ down to $-3$ .

Figure 10.4: Illustrating Bernoulli’s inequality for

r=3

10.4.3 AM–GM inequality

Given $a,b\geq 0$ , the AM-GM inequality relates the arithmetic mean (AM) $\frac{a+b}{2}$ with the geometric mean (GM) $\sqrt{ab}$ .

Theorem 10.23.

\sqrt{ab}\leq\frac{a+b}{2}

(10.5)

for all $a,b\geq 0$ .

Proof.

	$\displaystyle 0$	$\displaystyle\leq(a-b)^{2}$
		$\displaystyle=a^{2}-2ab+b^{2}$
		$\displaystyle=a^{2}+2ab+b^{2}-4ab$
		$\displaystyle=(a+b)^{2}-4ab.$

Hence

\sqrt{ab}\leq\frac{a+b}{2}

Exercise 10.24.

Prove that in (10.5) equality holds if and only if $a=b$ .

Solution (please try for yourself before looking)

The goal is to prove

\sqrt{ab}=\frac{a+b}{2}\,\Leftrightarrow\,a=b.

Assume $b=a$ , then $\frac{a+a}{2}=a=\sqrt{a^{2}}$ and so $\sqrt{ab}=\frac{a+b}{2}$ .

Assume $\sqrt{ab}=\frac{a+b}{2}$ .

		$\displaystyle\sqrt{ab}=\frac{a+b}{2}$
	$\displaystyle\Leftrightarrow\,$	$\displaystyle 4ab=(a+b)^{2}$
	$\displaystyle\Leftrightarrow\,$	$\displaystyle 0=(a-b)^{2}$
	$\displaystyle\Leftrightarrow\,$	$\displaystyle a=b$

Example 10.25.

We can apply (10.5) twice to $a_{1},a_{2},a_{3},a_{4}$ to prove

\left(a_{1}a_{2}a_{3}a_{4}\right)^{\frac{1}{4}}\leq\frac{\left(a_{1}a_{2}% \right)^{\frac{1}{2}}+\left(a_{3}a_{4}\right)^{\frac{1}{2}}}{2}\leq\frac{a_{1}% +a_{2}+a_{3}+a_{4}}{4}.

(10.6)

In more general form the AM–GM inequality states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean.

Theorem 10.26 (Generalised AM-GM Inequality).

For any list of $n$ non-negative real numbers $a_{1},\ a_{2},\ \ldots,\ a_{n}$ ,

\sqrt[n]{a_{1}\cdot a_{2}\cdots a_{n}}\leq\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}.

Cauchy’s original proof of the AM-GM inequality used a special kind of forward-backward induction. Let $P(n)$ be the statement

\sqrt[n]{a_{1}\cdot a_{2}\cdots a_{n}}\leq\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}.

The idea of the proof is as follows.

1.

Prove each of $P(2^{n})$ .
2.

If $n<2^{k}$ then prove $P(2^{k})\Rightarrow P(n)$

To prove $P(2^{n})$ use (10.6) repeatedly.

If $n<2^{k}$ , to show $P(2^{k})\Rightarrow P(n)$ let $A$ be defined as

A:=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}.

Consider the following product with $2^{k}$ terms:

\left(a_{1}a_{2}\cdots a_{n}A^{2^{k}-n}\right),

and then apply $P(2^{k})$ so

\left(a_{1}a_{2}\cdots a_{n}A^{2^{k}-n}\right)^{1/2^{k}}\leq\frac{a_{1}+a_{2}+% \cdots+a_{n}+(2^{k}-n)A}{2^{k}}=\frac{2^{k}A}{2^{K}}=A.

By clearing the powers of $A$ to the right-hand side, we have

\left(a_{1}a_{2}\cdots a_{n}\right)^{1/2^{k}}\leq A^{{n}/{2^{k}}}.

Finally, raising both sides to the power $2^{k}/n$ gives

\left(a_{1}a_{2}\cdots a_{n}\right)^{1/n}\leq A=\frac{a_{1}+a_{2}+\cdots+a_{n}% }{n}.

Exercise 10.27.

Let $c_{1},\ldots,c_{n}$ be positive real numbers. Prove that

\left(\sum_{k=1}^{n}c_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{c_{k}}\right)\geq n% ^{2}.

Solution (please try for yourself before looking)

Applying the AM-GM inequality to $c_{1},\ldots,c_{n}$

\frac{c_{1}+c_{2}+\cdots+c_{n}}{n}\geq\sqrt[n]{c_{1}\cdot c_{2}\cdots c_{n}}.

Applying the AM-GM inequality to $\frac{1}{c_{1}},\ldots,\frac{1}{c_{n}}$

\frac{1}{n}\left(\frac{1}{c_{1}}+\frac{1}{c_{2}}+\cdots+\frac{1}{c_{n}}\right)% \geq\sqrt[n]{\frac{1}{c_{1}}\cdot\frac{1}{c_{2}}\cdots\frac{1}{c_{n}}}=\frac{1% }{\sqrt[n]{c_{1}\cdot c_{2}\cdots c_{n}}}.

Multiplying these two inequalities together, and multiplying by $n^{2}$ gives

\left(\sum_{k=1}^{n}c_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{c_{k}}\right)\geq n% ^{2}.

10.4.4 Cauchy–Schwarz inequality

Theorem 10.28 (Cauchy–Schwarz Inequality).

Let $a_{1},\,a_{2},\cdots,a_{n}$ and $b_{1},\,b_{2},\cdots,b_{n}$ be two sequences (lists) of real numbers then

a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}\leq\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_% {n}^{2}}\sqrt{b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}}.

(10.7)

(10.7) is one of the most important inequalities in mathematics. Using sigma-notation this is written as

\sum_{k=1}^{n}a_{k}b_{k}\leq\sqrt{\sum_{k=1}^{n}a_{k}^{2}}\sqrt{\sum_{k=1}^{n}% b_{k}^{2}}.

We present two proofs of this inequality here.

Cauchy-Schwarz Inequality - Proof 1

Proof.

First note that $(x-y)^{2}=x^{2}-2xy+y^{2}\geq 0$ for all $x$ , $y$ . In particular, let $x=a_{1}b_{2}$ and $y=a_{2}b_{1}$ then

0\leq(a_{1}b_{2})^{2}-2(a_{1}b_{2})(a_{2}b_{1})+(a_{2}b_{1})^{2}.

Add $2a_{1}b_{1}a_{2}b_{2}$ , $(a_{1}b_{1})^{2}$ and $(a_{2}b_{2})^{2}$ to both sides

a_{1}^{2}b_{1}^{2}+2a_{1}b_{1}a_{2}b_{2}+a_{2}^{2}b_{2}^{2}\leq a_{1}^{2}b_{1}% ^{2}+a_{1}^{2}b_{2}^{2}+a_{2}^{2}b_{1}^{2}+a_{2}^{2}b_{2}^{2},

and factor both sides.

(a_{1}b_{1}+a_{2}b_{2})^{2}\leq(a_{1}^{2}+a_{2}^{2})(b_{1}^{2}+b_{2}^{2}).

(10.8)

This gives (10.7) for $n=2$ , and is the crucial step in proving (10.7) in general.

Let $P(n)$ be the statement (10.7), that is

a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}\leq\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_% {n}^{2}}\sqrt{b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}}.

Then, since $a_{1}b_{1}\leq\sqrt{a_{1}^{2}}\sqrt{b_{1}^{2}}$ we see that $P(1)$ is true for all $a_{1},b_{1}$ .

Note also, that by the algebra which proved (10.8) we see that $P(2)$ is also true.

Assume that $P(n)$ is true and consider

	$\displaystyle a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}+a_{n+1}b_{n+1}$
$\displaystyle=$	$\displaystyle(a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n})+a_{n+1}b_{n+1}$
$\displaystyle\leq$	$\displaystyle\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}\sqrt{b_{1}^{2}+b_{2}^% {2}+\cdots+b_{n}^{2}}+a_{n+1}b_{n+1}$	$\displaystyle\mbox{Applying }P(n)$
$\displaystyle\leq$	$\displaystyle\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}+a_{n+1}^{2}}\sqrt{b_{1% }^{2}+b_{2}^{2}+\cdots+a_{n}^{2}+a_{n+1}^{2}}$	$\displaystyle\mbox{Applying }P(2)$

Note, we proved $P(2)$ independently, and so may use it within this proof in the form.

\alpha\beta+a_{n+1}b_{n+1}\leq\sqrt{\alpha^{2}+a_{n+1}^{2}}\sqrt{\beta^{2}+b_{% n+1}^{2}}.

It therefore follows that $P(n+1)$ also holds, and so $P(n)$ follows by mathematical induction.

Cauchy-Schwarz Inequality - Proof 2

This proof of the Cauchy-Schwarz Inequality makes use of the following three lemmas.

Lemma 10.29.

The quadratic inequality $Ax^{2}+2Bx+C\geq 0$ holds for all $x\in\mathbb{R}$ if and only if $A>0$ and $AC-B^{2}\geq 0$ (or $A=B=0$ , and $C>0$ ).

Proof.

First suppose $A=B=0$ and $C\geq 0$ . Then

Ax^{2}+2Bx+C=C\geq 0.

Also, if $A>0$ and $AC-B^{2}\geq 0$ , then $C\geq\frac{B^{2}}{A}$ . Thus, by completing the square,

Ax^{2}+2Bx+C=A\left(x+\frac{B}{A}\right)^{2}-\frac{B^{2}}{A}+C\geq-\frac{B^{2}% }{A}+C\geq 0.

Hence, if $A>0$ and $AC-B^{2}\geq 0$ (or $A=B=0$ , and $C\geq 0$ ), we obtain $Ax^{2}+2Bx+C\geq 0$ .

For the other direction, suppose that $Ax^{2}+2Bx+C\geq 0$ holds for all $x\in\mathbb{R}$ .

If $A=0$ then $Ax^{2}+2Bx+C=2Bx+C\geq 0$ holds for all $x\in\mathbb{R}$ . This means that $B=0$ and $C\geq 0$ .

If $A\neq 0$ , then by completing the square, we have

Ax^{2}+2Bx+C=A\left(x+\frac{B}{A}\right)^{2}-\frac{B^{2}}{A}+C\geq 0,

for all $x\in\mathbb{R}$ . This means $A>0$ and its minimum value $-\frac{B^{2}}{A}+C$ must be non-negative i.e.

-\frac{B^{2}}{A}+C\geq 0\iff AC-B^{2}\geq 0.

Lemma 10.30.

Let $n\in\mathbb{N}$ and $a_{1},\ldots,a_{n},b_{1},\ldots,b_{n}\in\mathbb{R}$ . Then,

\sum_{k=1}^{n}(a_{k}x+b_{k})^{2}\geq 0,

for every $x\in\mathbb{R}$ .

Proof.

We prove this by induction on $n$ .

Let $P(n)$ be the statement “ $\sum_{i=1}^{n}(a_{i}x+b_{i})^{2}\geq 0$ ”.

For $n=1$ we have

(a_{1}x+b_{1})^{2}\geq 0

since $y^{2}\geq 0$ for all $y\in\mathbb{R}$ . Hence $P(1)$ is true.

Assume $P(k)$ is true. Then

\sum_{i=1}^{k+1}(a_{i}x+b_{i})^{2}=\left(\sum_{i=1}^{k}(a_{i}x+b_{i})^{2}% \right)+(a_{k+1}x+b_{k+1})^{2}.

The first term is non-negative by $P(k)$ , and we have $(a_{k+1}x+b_{k+1})^{2}\geq 0$ .

Hence $P(k+1)$ is true.

Since $P(1)$ holds and $P(k)\Rightarrow P(k+1)$ , we conclude $P(n)$ holds for all $n\in\mathbb{N}$ by induction.

Lemma 10.31.

Let $n\in\mathbb{N}$ and $a_{1},\ldots,a_{n},b_{1},\ldots,b_{n}\in\mathbb{R}$ .Then

\sum_{k=1}^{n}(a_{k}x+b_{k})^{2}=x^{2}\sum_{k=1}^{n}a_{k}^{2}+2x\sum_{k=1}^{n}% a_{k}b_{k}+\sum_{k=1}^{n}b_{k}^{2}

for all $x\in\mathbb{R}$ .

Proof.

We prove by induction on $n$ . Let $P(n)$ be the statement

\sum_{i=1}^{n}(a_{i}x+b_{i})^{2}=x^{2}\sum_{i=1}^{n}a_{i}^{2}+2x\sum_{i=1}^{n}% a_{i}b_{i}+\sum_{i=1}^{n}b_{i}^{2}.

For $n=1$ we have

(a_{1}x+b_{1})^{2}=x^{2}a_{1}^{2}+2xa_{1}b_{1}+b_{1}^{2}.

Hence $P(1)$ is true.

Assume $P(k)$ is true and consider:

	$\displaystyle\sum_{i=1}^{k+1}(a_{i}x+b_{i})^{2}$	$\displaystyle=\left(\sum_{i=1}^{k}(a_{i}x+b_{i})^{2}\right)+(a_{k+1}x+b_{k+1})% ^{2}$
		$\displaystyle=\left(x^{2}\sum_{i=1}^{k}a_{i}^{2}+2x\sum_{i=1}^{k}a_{i}b_{i}+% \sum_{i=1}^{k}b_{i}^{2}\right)+(a_{k+1}x+b_{k+1})^{2},\text{ by $P(k)$},$
		$\displaystyle=\left(x^{2}\sum_{i=1}^{k}a_{i}^{2}+2x\sum_{i=1}^{k}a_{i}b_{i}+% \sum_{i=1}^{k}b_{i}^{2}\right)$
		$\displaystyle\hphantom{=}+(x^{2}a_{k+1}^{2}+2xa_{k+1}b_{k+1}+b_{k+1}^{2})$
		$\displaystyle=x^{2}\left(\sum_{i=1}^{k}a_{i}^{2}+a_{k+1}^{2}\right)$
		$\displaystyle\hphantom{=}+2x\left(\sum_{i=1}^{k}a_{i}b_{i}+a_{k+1}b_{k+1}\right)$
		$\displaystyle\hphantom{=}+\left(\sum_{i=1}^{k}b_{i}^{2}+b_{k+1}^{2}\right)$
		$\displaystyle=x^{2}\sum_{i=1}^{k+1}a_{i}^{2}+2x\sum_{i=1}^{k+1}a_{i}b_{i}+\sum% _{i=1}^{k+1}b_{i}^{2}.$

Hence $P(k+1)$ is true.

Since $P(1)$ holds and $P(k)\Rightarrow P(k+1)$ , we conclude $P(n)$ holds for all $n\in\mathbb{N}$ by induction.

Another proof of the Cauchy-Schwarz Inequality is as follows.

Proof.

By Lemma 10.30 and Lemma 10.31, we see that

\sum_{k=1}^{n}(a_{k}x+b_{k})^{2}=x^{2}\sum_{k=1}^{n}a_{k}^{2}+2x\sum_{k=1}^{n}% a_{k}b_{k}+\sum_{k=1}^{n}b_{k}^{2}\geq 0.

By Lemma 10.29, we know $Ax^{2}+2Bx+C\geq 0$ for all $x\in\mathbb{R}$ if and only if $A>0$ and $AC-B^{2}\geq 0$ (or $A=B=0$ , and $C>0$ ). So take $A=\sum_{k=1}^{n}a_{k}^{2}\geq 0$ , $B=\sum_{k=1}^{n}a_{k}b_{k}$ and $C=\sum_{k=1}^{n}b_{k}^{2}$ .

If $A=B=0$ and $C>0$ , then $a_{k}=0$ for all $k=1,2,\ldots,n$ and the stated inequality holds since $0\leq 0$ .

If $A>0$ and $AC-B^{2}\geq 0$ , we conclude that

B^{2}\leq AC\iff\left(\sum_{k=1}^{n}a_{k}b_{k}\right)^{2}\leq\sum_{k=1}^{n}a_{% k}^{2}\sum_{k=1}^{n}b_{k}^{2}

and thus,

\sum_{k=1}^{n}a_{k}b_{k}\leq\sqrt{\sum_{k=1}^{n}a_{k}^{2}}\sqrt{\sum_{k=1}^{n}% b_{k}^{2}.}

Exercise 10.32.

Prove that for all $a_{1},a_{2}\cdots a_{n}$

\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\leq\sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a% _{n}^{2}}{n}}.

Solution (please try for yourself before looking)

Start by taking (10.7) with $b_{1}=b_{2}=\cdots=b_{n}=1$ .

a_{1}+a_{2}+\cdots+a_{n}\leq\sqrt{n}\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}.

Divide both sides by $n$ :

	$\displaystyle\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}$	$\displaystyle\leq\frac{\sqrt{n}}{n}\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}$
		$\displaystyle=\frac{1}{\sqrt{n}}\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}$
		$\displaystyle=\sqrt{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}}$

Exercise 10.33.

Let $c_{1},\ldots,c_{n}$ be positive real numbers. Prove that

\left(\sum_{k=1}^{n}c_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{c_{k}}\right)\geq n% ^{2}.

Solution (please try for yourself before looking)

Let $a_{k}=\sqrt{c_{k}}$ and $b_{k}=\frac{1}{\sqrt{c_{k}}}$ . Apply the Cauchy-Schwarz inequality

\sum_{k=1}^{n}a_{k}^{2}\sum_{k=1}^{n}b_{k}^{2}\geq\left(\sum_{k=1}^{n}a_{k}b_{% k}\right)^{2}.

Taking these values

\left(\sum_{k=1}^{n}c_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{c_{k}}\right)\geq% \left(\sum_{k=1}^{n}\sqrt{c_{k}}\frac{1}{\sqrt{c_{k}}}\right)^{2}=\left(\sum_{% k=1}^{n}1\right)^{2}=n^{2}.

Whenever we see a new inequality it is sensible to ask “how good is this?”. That is, does equality ever hold? To see when $a_{1}b_{1}+a_{2}b_{2}+\cdots+a_{n}b_{n}$ actually equals

\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}\sqrt{b_{1}^{2}+b_{2}^{2}+\cdots+b_% {n}^{2}}

we consider vectors in the plane.

10.4.5 The Cauchy–Schwarz Inequality and Vectors

Consider vectors $\vec{a}$ and $\vec{b}$ in $\mathbb{R}^{2}$ to be any two vectors in the plane.

\vec{a}=\begin{pmatrix}a_{1}\\ a_{2}\end{pmatrix},\quad\vec{b}=\begin{pmatrix}b_{1}\\ b_{2}\end{pmatrix}.

Then the length or norm of the vector is given by

||\vec{a}||=\sqrt{a_{1}^{2}+a_{2}^{2}}.

This follows from the Pythagorean Theorem. Notice that the term $\sqrt{a_{1}^{2}+a_{2}^{2}}$ appears as part of the right-hand side of (10.7). We can therefore consider how to interpret the left-hand side of (10.7) in terms of vectors $\vec{a}$ and $\vec{b}$ .

Let us assume that $\vec{a}\neq\vec{0}$ and $\vec{b}\neq\vec{0}$ . If we use our particular values to show when equality holds in (10.7), then we are trying to solve equation

(a_{1}b_{1}+a_{2}b_{2})^{2}=(a_{1}^{2}+a_{2}^{2})(b_{1}^{2}+b_{2}^{2}).

Looking back at the special case $P(2)$ of the Cauchy-Schwarz inequality, the algebra is identical:

		$\displaystyle(a_{1}b_{1}+a_{2}b_{2})^{2}=(a_{1}^{2}+a_{2}^{2})(b_{1}^{2}+b_{2}% ^{2})$
	$\displaystyle\Leftrightarrow\quad$	$\displaystyle 0=(a_{1}b_{2})^{2}-2(a_{1}b_{2})(a_{2}b_{1})+(a_{2}b_{1})^{2}.$
	$\displaystyle\Leftrightarrow\quad$	$\displaystyle 0=(a_{1}b_{2}-a_{2}b_{1})^{2}.$

This proves that equality holds in (10.7) if and only if $a_{1}b_{2}=a_{2}b_{1}$ .

The goal is to show $a_{1}b_{2}=a_{2}b_{1}$ if and only if $\vec{a}$ and $\vec{b}$ are parallel.

1.

Assume $a_{2}=0$ . Then since $\vec{a}\neq\vec{0}$ , we have $a_{1}\neq 0$ . Then $b_{2}=0$ and $a_{1}b_{2}=a_{2}b_{1}$ if and only if $\vec{a}$ and $\vec{b}$ are parallel.
2.

Assume $b_{2}=0$ . Then since $\vec{b}\neq\vec{0}$ , we have $b_{1}\neq 0$ . Then $a_{2}=0$ and $a_{1}b_{2}=a_{2}b_{1}$ if and only if $\vec{a}$ and $\vec{b}$ are parallel
3.

Assume $a_{2}\neq 0$ and $b_{2}\neq 0$ . Divide both sides by $a_{2}b_{2}$ to conclude that

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$

and hence $a_{1}b_{2}=a_{2}b_{1}$ if and only if $\vec{a}$ and $\vec{b}$ are parallel.

Hence, in all cases $a_{1}b_{2}=a_{2}b_{1}$ if and only if $\vec{a}$ and $\vec{b}$ are parallel.

Above we have worked directly in $\mathbb{R}^{2}$ . Much of the above generalised to vectors, and will be covered fully in a linear algebra course.

The inner product or dot product of two vectors

\vec{a}=\begin{pmatrix}a_{1}\\ a_{2}\end{pmatrix},\quad\vec{b}=\begin{pmatrix}b_{1}\\ b_{2}\end{pmatrix}.

is given by

\langle\vec{a},\vec{b}\rangle=\vec{a}\cdot\vec{b}=a_{1}b_{1}+a_{2}b_{2}.

Notice this is exactly the left-hand side of (10.7) for vectors in $\mathbb{R}^{2}$ . In $\mathbb{R}^{2}$ we can prove directly that

\langle\vec{a},\vec{b}\rangle=||\vec{a}||\,||\vec{b}||\cos{\theta}

where $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$ .

As we shall soon see in linear algebra courses, we can generalize both the length/norm of a vector and the inner product two of $n$ dimensional vectors.

||\vec{a}||:=\left(\sum_{k=1}^{n}a_{k}\right)^{\frac{1}{2}},\mbox{ and }% \langle\vec{a},\vec{b}\rangle:=\sum_{k=1}^{n}a_{k}b_{k}

and the Cauchy-Schwarz inequality shows that

\langle\vec{a},\vec{b}\rangle\leq||\vec{a}||\,||\vec{b}||

for any two vectors.

Indeed, mathematicians choose to generalise the idea of angle $\theta$ between two (non-zero) vectors in general by setting

\cos{\theta}:=\frac{\langle\vec{a},\vec{b}\rangle}{||\vec{a}||\,||\vec{b}||}.

If particular equality holds in the Cauchy-Schwarz inequality if and only if the vectors are parallel. This is one reason why the Cauchy-Schwarz inequality proves to be so important.