Skip to main content

Section 4.5 Some sample exercises

Show that if \(a,b \in\RR\text{,}\) then

\begin{equation*} -\max\{a^2,b^2\}\leq ab\leq \max\{a^2,b^2\}\text{.} \end{equation*}
Solution.

We may suppose \(|a|\leq |b|\text{.}\) Then

\begin{equation*} ab\leq |ab|=|a|\cdot|b|\leq |b|\cdot |b|=|b|^2=b^2= \max\{a^2,b^2\}\text{.} \end{equation*}

Similarly,

\begin{equation*} ab\geq -|ab|=-|a|\cdot |b| \geq -|b|^2=-b^2=- \max\{a^2,b^2\}\text{.} \end{equation*}

Show that for every \(a,b,c\in\RR\text{,}\) we have \(|a-c|\leq |a-b|+|b-c|\text{.}\)

Solution.

We add and subtract \(b\) from \(a-c\) and use the triangle inequality:

\begin{equation*} |a-c|=|a-b+b-c|\leq |a-b|+|b-c|\text{.} \end{equation*}

Show that if \(a,b \in \RR\text{,}\) then \(|a+b|\geq |a|-|b|\text{.}\) In fact, show that \(|a+b|\geq \left||a|-|b|\right|\text{.}\)

Solution.

By applying the triangle inequality \(|x+y|\leq |x|+|y|\) with \(x=a+b\) and \(y=-b\text{,}\)

\begin{equation*} |a|=|a+b-b|\leq |a+b|+|b| \end{equation*}

and subtracting \(|b|\) from both sides gives \(|a+b|\geq |a|-|b|\text{.}\) A similar proof shows that \(|a+b|\geq |b|-|a|=-(|a|-|b|)\text{.}\) Since \(||a|-|b||\) is either \(\pm (|a|-|b|)\) (depending on whether \(|a|-|b|\) is negative or not), this means \(|a+b|\geq \left||a|-|b|\right|\text{.}\)

Show that if \(0\lt a\leq b\text{,}\) then

\begin{equation*} a\leq \frac{a+b}{2}\leq b\text{.} \end{equation*}

Show that either of these inequalities is an equality if and only if \(a=b\text{.}\)

Solution.

Since \(a\leq b\text{,}\) we have \(a+b\leq b+b=2b\text{,}\) and so \(\frac{a+b}{2} \leq b\text{.}\) The other inequality has a similar proof. If either of the inequalities is an equality, we can just solve it to get \(a=b\text{.}\)

Show that if \(a,b,c,d>0\) and \(0\lt \frac{a}{b}\lt \frac{c}{d}\text{,}\) then

\begin{equation*} \frac{a}{b} \lt \frac{a+c}{b+d}\lt \frac{c}{d}\text{.} \end{equation*}
Solution.

Multiplying all sides by \(b+d\text{,}\) this string of inequalities is equivalent to

\begin{equation*} \frac{a}{b}(b+d) \lt a+c \lt \frac{c}{d}(b+d)\text{,} \end{equation*}

which implies

\begin{equation*} a+\frac{a}{b}d\lt a+c \lt \frac{d}{d}b+c\text{.} \end{equation*}

Since \(\frac{a}{b}\lt \frac{c}{d}\text{,}\) we know \(\frac{a}{b}d \lt \frac{c}{d}d=c\text{,}\) and adding \(a\) to both sides gives the first of the inequalities above. A similar argument gives the second inequality above.

Show the following for \(a,b,c,d\in\RR\text{:}\)

  1. \(\displaystyle (a^{2}-b^{2})(c^2-d^2)\leq (ac-bd)^2\)

  2. \((a^2+b^2)(c^2+d^2)\geq (ac+bd)^2\text{.}\)

  3. \((a^2-b^2)^2\geq 4ab(a-b)^2\text{.}\)

Solution.

  1. This inequality holds if and only if (after distributing the products)

    \begin{equation*} a^2c^2-a^2d^2-b^2c^2+b^2d^2\leq a^2c^2-2acbd+b^2d^2 \end{equation*}

    and, after cancelling common terms, this is equivalent to

    \begin{equation*} -a^2d^2-b^2c^2\leq -2acbd \end{equation*}

    which implies

    \begin{equation*} a^2d^2+b^2c^2\geq 2acbd\text{.} \end{equation*}

    Now we have a product of terms at most a sum, so this suggests using something like the AM-GM inequality. Indeed, the above inequality is true by the AM-GM inequality \(\sqrt{xy}\leq \frac{x+y}{2}\) with \(x=ad\) and \(y=bc\text{:}\)

    \begin{equation*} \frac{a^{2}d^{2}+b^2c^2}{2} \geq abcd \end{equation*}

  2. Again, by the AM-GM inequality

    \begin{align*} (ac+bd)^2 \amp = a^{2}c^{2}+2abcd+b^{2}d^{2}\\ \amp \leq a^{2}c^{2}+2\frac{a^2d^2+b^2c^2}{2}+b^{2}d^{2}\\ \amp =a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}\\ \amp = (ac+bd)^2\text{.} \end{align*}

  3. By the AM-GM inequality,

    \begin{equation*} (ab)^{\frac{1}{2}} \leq \frac{a+b}{2} \end{equation*}

    and so

    \begin{equation*} 4ab(a-b)^2 \leq 4 \left(\frac{a+b}{2}\right)^2(a-b)^2=(a+b)^2(a-b)^2=((a+b)(a-b))^2=(a^2-b^2)^2\text{.} \end{equation*}

Show that for \(a,b\in\RR\) and \(d>0\text{,}\)

\begin{equation*} ab \leq \frac{a^2}{d} + 2db^2\text{.} \end{equation*}

Hint: First \(ab=\frac{a}{c}(bc)\text{.}\)

Solution.

Notice that by Lemma 4.4.1 or the AM/GM inequality,

\begin{equation*} ab = \left(\frac{a}{c}\right) (bc) \leq \frac{\frac{a^2}{c^2}+b^2c^2}{2} \end{equation*}

Set \(c=\sqrt{2d}\text{,}\) then gives the claim.

Show that for \(a>0\text{,}\) \(a+\frac{1}{a}\geq 2\text{.}\)

Solution.

This follows from Theorem 4.4.3:

\begin{equation*} a+\frac{1}{a} \geq \sqrt{a\cdot \frac{1}{a}}=1\text{.} \end{equation*}

Given that \(x,y,z\in\mathbb{N}\text{,}\) solve

\begin{equation*} \left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)=3\text{.} \end{equation*}
Solution.

We claim that the only solutions \((x,y,z)\in \mathbb N\) with \(x\le y\le z\) are \((1,3,8)\text{,}\) \((1,4,5)\text{,}\) and \((2,2,3)\text{.}\)

We can assume \(0\lt x\leq y\leq z\text{.}\) We see that the biggest factor \(1+\frac1x\) is at least \(\sqrt[3]3\text{,}\) which implies \(x\leq 2\text{,}\) so \(x=1\) or \(2\text{.}\)

Case 1: \(x=1\text{.}\) Then the equation becomes

\begin{equation*} \left(1+\frac1y\right)\left(1+\frac1z\right)=\frac32 \end{equation*}

and so \(1+\frac1y\geq\sqrt{\frac32}\text{,}\) i.e. \(y\leq 4\text{.}\) Thus the allowed cases \(y=3\) and \(y=4\) can be easily checked by hand: \(y=3\) leads to \(1+\frac1z = \frac98\text{,}\) i.e. \(z=8\text{,}\) whereas \(y=4\) leads to \(1+\frac1z = \frac{6}{5}\text{,}\) i.e. \(z=5\text{.}\)

Case 2: \(x=2\text{.}\) Then the equation becomes

\begin{equation*} \left(1+\frac1y\right)\left(1+\frac1z\right)=2 \end{equation*}

and we conclude \(1+\frac1y\ge\sqrt{2}\text{,}\) i.e. \(y\leq 2\) and together with \(y\geq x\) this means \(y=2\) and finally \(1+\frac1z=\frac43\text{,}\) i.e. \(z=3\text{.}\)