# 3.2: More Direct Proofs

- Page ID
- 19375

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In creating a direct proof, we need to look at our hypotheses, consider the desired conclusion, and develop a strategy for transforming \(\text{A}\) into \(\text{B}\). Quite often you’ll find it easy to make several deductions from the hypotheses, but none of them seems to be headed in the direction of the desired conclusion. The usual advice at this stage is “Try working backwards from the conclusion.”^{1}

There is a lovely result known as the “arithmetic-geometric mean inequality” whose proof epitomizes this approach. Basically this inequality compares two different ways of getting an “average” between two real numbers. The arithmetic mean of two real numbers \(a\) and \(b\) is the one you’re probably used to, \(\dfrac{(a+b)}{2}\). Many people just call this the “mean” of \(a\) and \(b\) without using the modifier “arithmetic” but as we’ll see, our notion of what intermediate value to use in between two numbers is dependent on context. Consider the following two sequences of numbers (both of which have a missing entry)

\(2\; 9\;16\; 23\; \underline{\;\;} \;37\; 44\)

and

\(3 \;6\; 12\; 24\; \underline{\;\;} \;96\; 192.\)

How should we fill in the blanks?

The first sequence is an **arithmetic sequence**. Arithmetic sequences are characterized by the property that the difference between successive terms is a constant. The second sequence is a **geometric sequence**. Geometric sequences have the property that the ratio of successive terms is a constant. The blank in the first sequence should be filled with the arithmetic mean of the surrounding entries \(\dfrac{(23 + 37)}{2} = 30\). The blank in the second sequence should be filled using the geometric mean of its surrounding entries: \(\sqrt{24 · 96} = 48\).

Given that we accept the utility of having two inequivalent concepts of **mean **that can be used in different contexts, it is interesting to see how these two means compare to one another. The arithmetic-geometric mean inequality states that the arithmetic mean is always bigger.

\[∀a, b ∈ \mathbb{R}, a, b ≥ 0 \implies \dfrac{a + b}{2} ≥ \sqrt{ab}\]

In proving this statement we have little choice but to work backwards from the conclusion because the only hypothesis we have to work with is that \(a\) and \(b\) are non-negative real numbers – which isn’t a particularly potent tool. But what should we do? There isn’t a good response to that question, we’ll just have to try a bunch of different things and hope that something will work out. When we finally get around to writing up our proof though, we’ll have to rearrange the statements in the opposite order from the way they were discovered. This means that we would be ill-advised to make any uni-directional inferences, we should strive to make biconditional connections between our statements (or else try to intentionally make converse errors).

The first thing that appeals to your humble author is to eliminate both the fractions and the radicals. . .

\[\dfrac{a + b}{2} ≥ \sqrt{ab} \\ \iff a + b ≥ 2 \sqrt{ab} \\ \iff (a + b)^2 ≥ 4ab \\ \iff a^2 + 2ab + b^2 ≥ 4ab \]

One of the steps above involves squaring both sides of an inequality. We need to ask ourselves if this step is really reversible. In other words, is the following conditional true?

\[∀x, y ∈ \mathbb{R}^{\text{noneg}}, x ≥ y \implies \sqrt{x} ≥ \sqrt{y}\]

Provide a justification for the previous implication.

What should we try next? There’s really no good justification for this but experience working with quadratic polynomials either in equalities or inequalities leads most people to try “moving everything to one side,” that is, manipulating things so that one side of the equation or inequality is zero.

\[a^2 + 2ab + b^2 ≥ 4ab \\ \iff a^2 − 2ab + b^2 ≥ 0 \]

Whoa! We’re done! Do you see why? If not, I’ll give you one hint: the square of any real number is greater than or equal to zero.

Re-assemble all of the steps taken in the previous few paragraphs into a proof of the arithmetic-geometric mean inequality.

## Exercises:

Suppose you have a savings account which bears interest compounded monthly. The July statement shows a balance of \($2104.87\) and the September statement shows a balance \($2125.97\). What would be the balance on the (missing) August statement?

Recall that a quadratic equation \(ax^2 +bx+c = 0\) has two real solutions if and only if the discriminant \(b^2 −4ac\) is positive. Prove that if a and c have different signs then the quadratic equation has two real solutions.

Prove that if \(x^3 − x^2\) is negative then \(3x + 4 < 7\).

Prove that for all integers \(a\), \(b\), and \(c\), if \(a|b\) and \(a|(b + c)\), then \(a|c\).

Show that if \(x\) is a positive real number, then \(x + \dfrac{1}{x} ≥ 2\).

Prove that for all real numbers \(a\), \(b\), and \(c\), if \(ac < 0\), then the quadratic equation \(ax^2 + bx + c = 0\) has two real solutions.

**Hint**-
The quadratic equation \(ax^2 + bx + c = 0\) has two real solutions if and only if \(b^2 − 4ac > 0\) and \(a \neq 0\).

Show that \(\binom{n}{k} · \binom{k}{r} = \binom{n}{r} · \binom{n−r}{k−r}\) (for all integers \(r\), \(k\) and \(n\) with \(r ≤ k ≤ n\).

In proving the **product rule** in Calculus using the definition of the derivative, we might start our proof with:

\(\dfrac{d}{dx} (f(x) · g(x)) \\ \lim_{h \rightarrow 0} \dfrac{f(x + h) · g(x + h) − f(x) · g(x)}{h} \)

The last two lines of our proof should be:

\(\lim_{h→0} \dfrac{f(x + h) − f(x)}{h} · g(x) + f(x) · \lim_{h→0} \dfrac{g(x + h) − g(x)}{h} \\ = \dfrac{d}{dx} (f(x)) · g(x) + f(x) · \dfrac{d}{dx} (g(x))\)

Fill in the rest of the proof.