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.

Answer:

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. $a^3+1+1+6≥3\sqrt[3]{a^3·1·1}+6$

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 [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.

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

[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.

Answer:

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 $0\le a\le 4$.

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. [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]

Answer:

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]

Therefore A is the answer.

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]

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