## Tuesday, January 19, 2016

### 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 [MATH]\sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \frac{4}{9}[/MATH].

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.

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

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. $a^3+1+1+6≥3\sqrt[3]{a^3·1·1}+6$

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 [MATH]\sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le\frac{4}{9}[/MATH], and you're then told to use the extended Cauchy-Schwarz inequality to finish the proof.

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

A. To cross-multiply both sides and get [MATH]\frac{9}{4}\le \sum_{\text{cyclic}}^{} \frac{3(a+2)}{a}[/MATH].
B. To add [MATH]\sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}[/MATH] 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.

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

[MATH]\sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

Extended Cauchy-Schwarz inequality tells us

[MATH]\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}[/MATH]

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

[MATH]\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)}[/MATH]

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

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

[MATH]\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)}[/MATH]

[MATH]\sum_{\text{cyclic}}^{}\frac{a+2}{3(a+2)}\le \frac{4}{9}+\sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}[/MATH]

[MATH]\frac{4}{3}- \frac{4}{9}\le \sum_{\text{cyclic}}^{}\frac{2}{3(a+2)}[/MATH]

[MATH]\frac{4}{3}\le \sum_{\text{cyclic}}^{}\frac{1}{a+2}[/MATH]

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

[MATH]\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*}[/MATH]

And we're hence done.

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

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

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

[MATH] \frac{n^2}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}\le a_1+a_2+\cdots+a_n[/MATH]

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

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

[MATH]\frac{\frac{1}{a}+\frac{1}{2}}{4}\ge \frac{1}{a+2}[/MATH]

[MATH]\frac{\frac{a}{a}+\frac{a}{2}}{4}\ge \frac{a}{a+2}[/MATH]

[MATH]\frac{1+\frac{a}{2}}{4(3)}\ge \frac{a}{3(a+2)}[/MATH]

[MATH]\sum_{\text{cyclic}}^{}\frac{1+\frac{a}{2}}{4(3)}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\frac{4}{12}+\frac{a+b+c+d}{24}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\frac{4}{12}+\frac{4}{24}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\frac{1}{2}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le \frac{1}{2}[/MATH]

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

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

[MATH]a+1+1≥\frac{9}{\frac{1}{a}+1+1}[/MATH]

[MATH]\frac{\frac{1}{a}+1+1}{9}\ge \frac{1}{a+2}[/MATH]

[MATH]\frac{\frac{1}{a}+2}{9}\ge \frac{1}{a+2}[/MATH]

[MATH]\frac{\frac{a}{a}+2a}{9}\ge \frac{a}{a+2}[/MATH]

[MATH]\frac{1+2a}{9(3)}\ge \frac{a}{3(a+2)}[/MATH]

[MATH]\sum_{\text{cyclic}}^{}\frac{1+2a}{27}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\frac{4}{27}+\frac{2(a+b+c+d)}{27}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\frac{4}{27}+\frac{2(4)}{27}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\frac{4}{9}\ge \sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}[/MATH]

[MATH]\sum_{\text{cyclic}}^{}\frac{a}{3(a+2)}\le \frac{4}{9}[/MATH]

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 [MATH]f(a)=\frac{a}{a^3+8}[/MATH]?

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 $0\le a\le 4$.

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. [MATH]\frac{a}{a^3+8}\le \frac{2a+1}{27}[/MATH]
B. [MATH]\frac{a}{a^3+8}\le \frac{2a-1}{27}[/MATH]
C. [MATH]\frac{a}{a^3+8}\le \frac{a+2}{27}[/MATH]
D. [MATH]\frac{a}{a^3+8}\le \frac{a-2}{27}[/MATH]

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

[MATH]f'(k)>\frac{f(a)-f(k)}{a-k}[/MATH]

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

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

[MATH]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}[/MATH]

Therefore, we can set up the following:

[MATH]\frac{2}{27}>\frac{\frac{a}{a^3+8}-\frac{1}{9}}{a-1}[/MATH]

[MATH]\frac{2(a-1)}{27}>\frac{a}{a^3+8}-\frac{1}{9}[/MATH]

[MATH]\frac{2a+1}{27}>\frac{a}{a^3+8}[/MATH]

If we proceed, we see that we get:

[MATH]\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*}[/MATH]