9.2 Workshop 2 (Reading)

9.2.1 Preparatory Exercises (30 mins)

Task 9.11.

Without using a calculator, try and prove that

\sqrt{17}<1+\sqrt{10}

using:

(a)

an algebraic argument.
(b)

a geometric (visual) argument.

Come to the Reading workshop prepared to share and discuss your arguments with your group.

9.2.2 B5 Reading Workshop Supplementary Material

For Task B5.2.1:

Theorem (AM-GM inequality).

\frac{a+b}{2}\geq\sqrt{ab}\quad\mbox{for all}\quad a,b\geq 0.

Consider the following three proofs of the AM-GM Inequality.

Geometric proof 1: This is a Proof without words.

(a+b)^{2}-(a-b)^{2}=4ab

\implies\frac{a+b}{2}\geq\sqrt{ab}.

$\square$

Algebraic proof 1: Observe that

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

Rearranging the last line, we get

\frac{a+b}{2}\geq\sqrt{ab}.

$\square$

Geometric proof 2: This is another Proof without words.

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

$\square$

For Task B5.2.2:

Theorem (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: This is the first proof of the AM-GM inequality. It uses a special kind of forward-backward induction.

Proof.

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 $P(2^{k})$ for each $k\in\mathbb{N}$ .
2.

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

To prove $P(2^{k})$ for each integer $k>1$ , we apply the simple AM-GM inequality repeatedly using an induction argument.

The statement $P(2)$ holds immediately from the simple AM-GM inequality. Assuming $P(2^{k-1})$ holds, we have

	$\displaystyle\left(a_{1}a_{2}\cdots a_{2^{k}-1}a_{2^{k}}\right)^{\frac{1}{2^{k% }}}$	$\displaystyle=\left((a_{1}a_{2})^{\frac{1}{2}}\cdots(a_{2^{k}-1}a_{2^{k}})^{% \frac{1}{2}}\right)^{\frac{1}{2^{k-1}}}$
		$\displaystyle\leq\frac{\left(a_{1}a_{2}\right)^{\frac{1}{2}}+\cdots+\left(a_{2% ^{k}-1}a_{2^{k}}\right)^{\frac{1}{2}}}{2^{k-1}}$
		$\displaystyle\leq\frac{a_{1}+a_{2}+\cdots+a_{2^{k}-1}+a_{2^{k}}}{2^{k}}.$

So by induction on $k$ , $P(2^{k})$ holds for all $k\in\mathbb{N}$ .

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}.

Polya’s proof: George Polya used calculus to prove the AM-GM inequality.

Proof.

Let $f(x)=e^{x-1}-x$ (a sketch of the graph of $f$ is given below). Using basic calculus we see that

f^{\prime}(x)=e^{x-1}-1,\quad f^{\prime\prime}(x)=e^{x-1}.

Then $f(1)=0$ , $f^{\prime}(1)=0$ and $f^{\prime\prime}(x)>0$ for all $x$ .

Hence

x\leq e^{x-1}

(9.1)

with equality only when $x=1$ . (A sketch of the graphs of $e^{x-1}$ in blue and $x$ in red are given below).

Consider the non-negative real numbers $a_{1},\ a_{2},\ \ldots,\ a_{n}$ . If they are all zero then the AM-GM inequality holds.

Assume they are not all zero and define the mean $a$ to be

a=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}>0

By repeated application of (9.1), we see that

	$\displaystyle\frac{a_{1}}{a}\frac{a_{2}}{a}\cdots\frac{a_{n}}{a}$	$\displaystyle\leq e^{\frac{a_{1}}{a}-1}e^{\frac{a_{2}}{a}-1}\cdots e^{\frac{a_% {n}}{a}-1}$
		$\displaystyle=e^{\frac{a_{1}}{a}-1+\frac{a_{2}}{a}-1+\cdots+\frac{a_{n}}{a}-1}$
		$\displaystyle=e^{\frac{a_{1}+a_{2}+\cdots+a_{n}}{a}-n}$
		$\displaystyle=e^{\frac{na}{a}-n}$
		$\displaystyle=e^{0}=1.$

Equality holds on the first line if, and only if, $a_{k}=a$ for all $k=1,2,\ldots,n$ (i.e. $x=1$ in (9.1)).

Hence

\frac{a_{1}}{a}\frac{a_{2}}{a}\cdots\frac{a_{n}}{a}\leq 1

which implies that

a_{1}a_{2}\cdots a_{n}\leq a^{n}.

Hence,

\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\leq a=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}

which proves the general AM-GM inequality.

For Task B5.2.3:

Lemma (LEMMA A).

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 $C\geq 0$ ).

Proof.

This is almost identical to the problem in the B5 Exploration preparatory exercises.

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={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0% }\ldots}

[Complete this step.]

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 $C\geq 0$ . [Why?]

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

Ax^{2}+2Bx+C={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0% }\ldots},

[Complete this step.] 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 (LEMMA B).

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 [Complete the induction step.]

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 (LEMMA C).

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 $P(k)$},$
		$\displaystyle={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{% 1,0,0}\ldots}$
		$\displaystyle=x^{2}\sum_{i=1}^{k+1}a_{i}^{2}+2x\sum_{i=1}^{k+1}a_{i}b_{i}+\sum% _{i=1}^{k+1}b_{i}^{2}.$

[Complete the induction step.] 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.

Theorem (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

\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}}.

Proof.

By Lemma B and C, we see that

\sum_{k=1}^{n}(a_{k}x+b_{k})^{2}={\color[rgb]{1,0,0}\definecolor[named]{% pgfstrokecolor}{rgb}{1,0,0}\ldots}

[Complete this step.] By Lemma A, 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 $A={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\ldots}$ , $B={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\ldots}$ and $C={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\ldots}$ .

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 ….

[Complete this step.]

9.2.3 Workshop Tasks

Task 9.12 (Warm Up (10 mins)).

Share your proofs that

\sqrt{17}<1+\sqrt{10}

from the preparatory exercises with the rest of your group.

1.

Which proof is the most rigorous?
2.

Which proof is the simplest?
3.

Which proof do you prefer and why?

Figure 9.1: Visual proof of

\sqrt{17}<1+\sqrt{10}

9.2.4 The AM-GM Inequality

In its simplest form the AM–GM inequality is as follows.

Theorem (AM-GM inequality).

\frac{a+b}{2}\geq\sqrt{ab}\quad\mbox{for all}\quad a,b\geq 0.

(9.2)

This inequality relates the arithmetic mean $\frac{a+b}{2}$ with the geometric mean $\sqrt{ab}$ .

Task 9.13 (15 mins).

In your supplementary material, you’ve been given three proofs of the AM-GM inequality.

1.

Discuss these proofs with your group and clarify any steps you don’t understand.
2.

Which is the most rigorous?
3.

Which is the simplest?
4.

Which gives you the best insight into what the inequality is saying and why it is true?

EXTENSION: Can you use the AM-GM inequality to derive a lower bound for the arithmetic mean of more than two real numbers?

Consider the following three proofs

Geometric proof 1: This is a Proof without words.

(a+b)^{2}-(a-b)^{2}=4ab

\implies\frac{a+b}{2}\geq\sqrt{ab}.

$\square$

Algebraic proof 1: Observe that

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

Rearranging the last line, we get

\frac{a+b}{2}\geq\sqrt{ab}.

$\square$

Geometric proof 2: This is another Proof without words.

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

$\square$

Task 9.14 (20 mins).

In your supplementary material, you’ve been given two proofs of the generalised AM-GM inequality.

1.

Discuss these proofs with your group and clarify any steps you don’t understand.
2.

Which is the most rigorous?
3.

Which is the simplest?
4.

Cauchy only uses elementary algebra, but Polya uses some calculus. What are the relative merits of these approaches?

EXTENSION: Are there any corollaries you can pull from these proofs?

Theorem (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: This is the first proof of the AM-GM inequality. It uses a special kind of forward-backward induction.

Proof.

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 $P(2^{k})$ for each $k\in\mathbb{N}$ .
2.

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

To prove $P(2^{k})$ for each integer $k>1$ , we apply the simple AM-GM inequality repeatedly using an induction argument.

The statement $P(2)$ holds immediately from the simple AM-GM inequality. Assuming $P(2^{k-1})$ holds, we have

	$\displaystyle\left(a_{1}a_{2}\cdots a_{2^{k}-1}a_{2^{k}}\right)^{\frac{1}{2^{k% }}}$	$\displaystyle=\left((a_{1}a_{2})^{\frac{1}{2}}\cdots(a_{2^{k}-1}a_{2^{k}})^{% \frac{1}{2}}\right)^{\frac{1}{2^{k-1}}}$
		$\displaystyle\leq\frac{\left(a_{1}a_{2}\right)^{\frac{1}{2}}+\cdots+\left(a_{2% ^{k}-1}a_{2^{k}}\right)^{\frac{1}{2}}}{2^{k-1}}$
		$\displaystyle\leq\frac{a_{1}+a_{2}+\cdots+a_{2^{k}-1}+a_{2^{k}}}{2^{k}}.$

So by induction on $k$ , $P(2^{k})$ holds for all $k\in\mathbb{N}$ .

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}.

Polya’s proof: George Polya used calculus to prove the AM-GM inequality.

Proof.

Let $f(x)=e^{x-1}-x$ (a sketch of the graph of $f$ is given below). Using basic calculus we see that

f^{\prime}(x)=e^{x-1}-1,\quad f^{\prime\prime}(x)=e^{x-1}.

Then $f(1)=0$ , $f^{\prime}(1)=0$ and $f^{\prime\prime}(x)>0$ for all $x$ .

Hence

x\leq e^{x-1}

(9.3)

with equality only when $x=1$ . (A sketch of the graphs of $e^{x-1}$ in blue and $x$ in red are given below).

Consider the non-negative real numbers $a_{1},\ a_{2},\ \ldots,\ a_{n}$ . If they are all zero then the AM-GM inequality holds.

Assume they are not all zero and define the mean $a$ to be

a=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}>0.

By repeated application of (9.3), we see that

	$\displaystyle\frac{a_{1}}{a}\frac{a_{2}}{a}\cdots\frac{a_{n}}{a}$	$\displaystyle\leq e^{\frac{a_{1}}{a}-1}e^{\frac{a_{1}}{a}-1}\cdots e^{\frac{a_% {n}}{a}-1}$
		$\displaystyle=e^{\frac{a_{1}}{a}-1+\frac{a_{1}}{a}-1+\cdots+\frac{a_{n}}{a}-1}$
		$\displaystyle=e^{\frac{a_{1}+a_{2}+\cdots+a_{n}}{a}-n}$
		$\displaystyle=e^{\frac{na}{a}-n}$
		$\displaystyle=e^{0}=1.$

Equality holds on the first line if, and only if, $a_{k}=a$ for all $k=1,2,\ldots,n$ (i.e. $x=1$ in (9.3)).

Hence

\frac{a_{1}}{a}\frac{a_{2}}{a}\cdots\frac{a_{n}}{a}\leq 1

which implies that

a_{1}a_{2}\cdots a_{n}\leq a^{n}.

Hence,

\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\leq a=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}

which proves the general AM-GM inequality.

Task 9.15 (20 mins).

In your supplementary material, you have a sparse outline of a proof of the Cauchy-Schwarz inequality and proofs of the results that it uses.

1.

Work with your group to complete the details of these proofs.
2.

Work together to clarify any steps that you don’t understand.

EXTENSION: Try and prove that the Cauchy-Schwarz inequality holds with equality if and only if there exists some $r\in\mathbb{R}$ such that $a_{i}=rb_{i}$ for all $i=1,2,\ldots,n$ .

Lemma (LEMMA A).

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 $C\geq 0$ ).

Proof.

This is almost identical to the problem in the B5 Exploration preparatory exercises.

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={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0% }\ldots}

[Complete this step.]

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 $C\geq 0$ . [Why?]

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

Ax^{2}+2Bx+C={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0% }\ldots},

[Complete this step.] 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 (LEMMA B).

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 [Complete the induction step.]

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 (LEMMA C).

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 $P(k)$},$
		$\displaystyle={\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{% 1,0,0}\ldots}$
		$\displaystyle=x^{2}\sum_{i=1}^{k+1}a_{i}^{2}+2x\sum_{i=1}^{k+1}a_{i}b_{i}+\sum% _{i=1}^{k+1}b_{i}^{2}.$

[Complete the induction step.] 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.

Theorem (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

\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}}.

Proof.

By Lemma B and C, we see that

\sum_{k=1}^{n}(a_{k}x+b_{k})^{2}={\color[rgb]{1,0,0}\definecolor[named]{% pgfstrokecolor}{rgb}{1,0,0}\ldots}

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 ….

[Complete this step.]

Task 9.16 (Optional Extension).

Use the AM-GM inequality to show that

3abc\leq a^{3}+b^{3}+c^{3}

for all $a,b,c\in\mathbb{R}$ .