Analysis Quiz 18: Multiple-Choice Test (Develop Problem Solving Skill)

Please answer the following questions based on the proving of the inequality below:
Given $a,\,b,\,c$ and $d$ are all positive real numbers such that $a+b+c+d=4$. Prove that $\displaystyle \sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \frac{4}{9}$.

Question 1: Do you think we can stick to the approach we employed in the previous quiz (17) for proving this inequality?
A. Yes.
B. No.

Answer:

No, we most definitely cannot because the previous approaches are all led to proving a weaker bound, i.e. $\displaystyle \sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \frac{12}{25}$, and $\displaystyle \frac{4}{9}\lt \frac{12}{25}$.

Question 2: If you're asked to use AM-GM inequality to deal with the expression $a^3+8$, which of the following way would you approach?
A. $a^3+8≥2\sqrt{a^3·8}$
B. $a^3+1+1+1+1+1+1+1+1≥9(a^3·1·1·1·1·1·1·1·1)^{\frac{1}{9}}$
C. $a^3+2+2+2+2≥5(a^3·2·2·2·2)^{\frac{1}{5}}$
D. $a^3+1+1+6≥3\sqrt[3]{a^3·1·1}+6$

Answer:

We need to work with a nicer and clean term, and out of all the options, it's not hard to see that only option D gives us the result $a^3+1+1+6≥3a+6=3(a+2)$.

We're hence sorely tempted to pick D as the answer, which we will confirm later if that is the right choice when we go through the subsequent questions.

Question 3: If you've made progress and got $\displaystyle \sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le\frac{4}{9}$, and you're then told to use the extended Cauchy-Schwarz inequality to finish the proof.

What would you do to the $\displaystyle \sum_{\text{cyclic}}^{} \frac{a}{3(a+2)}\le\frac{4}{9}$ in order to proceed?

A. To cross-multiply both sides and get $\displaystyle \frac{9}{4}\le \sum_{\text{cyclic}}^{} \frac{3(a+2)}{a}$.
B. To add $\displaystyle \sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}$ to both sides of the inequality.
C. We could immediately employ the formula without doing anything to the expression on the left of the inequality.

Answer:

From the first question, we know if we used the AM-GM inequality to get $a^3+1+1+6≥3a+6=3(a+2)$, then by rearranging it yields

$\displaystyle \sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

Extended Cauchy-Schwarz inequality tells us

$\displaystyle \frac{(a+b+c+d)^2}{x+y+z+t}\le \frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}+\frac{d^2}{t}$

We know if we straightaway apply the formula, it won't do us any good:

$\displaystyle \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2}{3(a+b+c+d)+4(3)(2)}$

You won't want to deal with $\displaystyle \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2}{3(a+b+c+d)+4(3)(2)}$.

But if we're to add $\displaystyle \sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}$ to both sides of the inequality, then we can see the light at the end of the tunnel.

$\displaystyle \sum_{\text{cyclic}}^{}\left(\frac{a}{3(a+2)}+\frac{2}{3(a+2)}\right)\le \frac{4}{9}+\sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}$

$\displaystyle \sum_{\text{cyclic}}^{}\frac{a+2}{3(a+2)}\le \frac{4}{9}+\sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}$

$\displaystyle \frac{4}{3}- \frac{4}{9}\le \sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}$

$\displaystyle \frac{4}{3}\le \sum_{\text{cyclic}}^{}\frac{1}{a+2}$

Now, we can safely apply the extended Cauchy-Schwarz inequality to $\displaystyle \sum_{\text{cyclic}}^{}\frac{1}{a+2}$:

$\displaystyle \begin{align*}\sum_{\text{cyclic}}^{}\frac{1}{a+2}&=\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}+\frac{1}{d+2}\\&\ge \frac{(1+1+1+1)^2}{a+b+c+d+4(2)}\\& \ge \frac{(4)^2}{4+4(2)}\\& \ge \frac{4}{1+2}\\& \ge \frac{4}{3}\end{align*}$

And we're hence done.

Question 4: Repeat question $3$ by using the AM-HM inequality to prove the inequality correct.

A. Use $\displaystyle a+2\ge \frac{4}{\frac{1}{a}+\frac{1}{2}}$.
B. Use $\displaystyle a+\frac{2}{3}+\frac{2}{3}+\frac{2}{3}\ge \frac{16}{\frac{1}{a}+\frac{3}{2}+\frac{3}{2}+\frac{3}{2}+\frac{3}{2}}$.
C. Use $\displaystyle a+1+1≥\frac{9}{\frac{1}{a}+1+1}$.
D. The AM-HM inequality can't be used to prove the given inequality correct.

Answer:

We're asked to use AM-HM inequality to prove $\displaystyle \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le\frac{4}{9}$, and AM-HM inequality tells us:

$\displaystyle \frac{n^2}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}\le a_1+a_2+\cdots+a_n$

Therefore, our focus should turn to manipulating $\displaystyle \frac{1}{a+2}\le \text{something}$, and we should investigate what that something is in the next step.

If we look at the first option, i.e. to use $\displaystyle a+2\ge \frac{4}{\frac{1}{a}+\frac{1}{2}}$, we end up with

$\displaystyle \frac{\frac{1}{a}+\frac{1}{2}}{4}\ge \frac{1}{a+2}$

$\displaystyle \frac{\frac{a}{a}+\frac{a}{2}}{4}\ge \frac{a}{a+2}$

$\displaystyle \frac{1+\frac{a}{2}}{4(3)}\ge \frac{a}{3(a+2)}$

$\displaystyle \sum_{\text{cyclic}}^{}\frac{1+\frac{a}{2}}{4(3)}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \frac{4}{12}+\frac{a+b+c+d}{24}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \frac{4}{12}+\frac{4}{24}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \frac{1}{2}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le \frac{1}{2}$

But $\displaystyle \frac{1}{2}\gt \frac{4}{9}$, therefore we can't conclude that $\displaystyle \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le \frac{4}{9}$ and A is out of the question.

By the similar token, it's not hard to see option C is the correct answer since:

$\displaystyle a+1+1≥\frac{9}{\frac{1}{a}+1+1}$

$\displaystyle \frac{\frac{1}{a}+1+1}{9}\ge \frac{1}{a+2}$

$\displaystyle \frac{\frac{1}{a}+2}{9}\ge \frac{1}{a+2}$

$\displaystyle \frac{\frac{a}{a}+2a}{9}\ge \frac{a}{a+2}$

$\displaystyle \frac{1+2a}{9(3)}\ge \frac{a}{3(a+2)}$

$\displaystyle \sum_{\text{cyclic}}^{}\frac{1+2a}{27}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \frac{4}{27}+\frac{2(a+b+c+d)}{27}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \frac{4}{27}+\frac{2(4)}{27}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \frac{4}{9}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}$

$\displaystyle \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le \frac{4}{9}$

And we're hence done and C is the correct answer.

Question 5: If you're to use the tangent line method to prove the inequality, what can you say of the concavity of the function of $\displaystyle f(a)=\frac{a}{a^3+8}$?

A. The function of $f(a)$ is a concave function for $0\le a\le 4$.
B. The function of $f(a)$ is a convex function for $0\le a\le 4$.

Answer:

If we use the graphical calculator online, for example wolfram alpha or the desmos graphing calculator, we will see the function of $f(a)$ is a concave function.

Question 6: If you're to use the tangent line method to prove the inequality, which of the following step would you have got?
A. $\displaystyle \frac{a}{a^3+8}\le \frac{2a+1}{27}$
B. $\displaystyle \frac{a}{a^3+8}\le \frac{2a-1}{27}$
C. $\displaystyle \frac{a}{a^3+8}\le \frac{a+2}{27}$
D. $\displaystyle \frac{a}{a^3+8}\le \frac{a-2}{27}$

Answer:

If we have a concave function, then by the theorem of tangent line method, we have

$\displaystyle f'(k)>\frac{f(a)-f(k)}{a-k}$

We normally would pick $k=1$ (by considering $a+b+c+d=4$) and that gives

$f(k)=f(1)=\dfrac{1}{1+8}=\dfrac{1}{9}$

$\displaystyle f'(k)=f'(1)=\frac{(a^3+8)(1)-(a)(3a^2)}{(a^3+8)^2}=\frac{(1+8)-3}{(1+8)^2}=\frac{6}{9(9)}=\frac{2}{27}$

Therefore, we can set up the following:

$\displaystyle \frac{2}{27}>\frac{\frac{a}{a^3+8}-\frac{1}{9}}{a-1}$

$\displaystyle \frac{2(a-1)}{27}>\frac{a}{a^3+8}-\frac{1}{9}$

$\displaystyle \frac{2a+1}{27}>\frac{a}{a^3+8}$

Therefore A is the answer.

If we proceed, we see that we get:

$\displaystyle \begin{align*}\sum_{\text{cyclic}}^{} \frac{a}{a^3+8}&\le \sum_{\text{cyclic}}^{} \frac{2a+1}{27}\\&\le \frac{2(a+b+c+d)+1(4)}{27}\\&\le \frac{2(4)+1(4)}{27}\\&\le \frac{4}{9}\end{align*}$