## Sunday, January 3, 2016

### Analysis Quiz 17: Multiple-Choice Test (Improve Logical Thinking and Develop Discerning Patterns Skills)

Question 1: Given $a,\,b,\,c$ and $d$ are all positive real numbers such that $a+b+c+d=4$. Prove that [MATH]\sum_{cyclic} \frac{a}{a^3+8}\lt \frac{12}{25}[/MATH].

Do you think you would need the given equality to prove for the inequality?

A. Yes.
B. No.

Of course we need the given equality to prove for the target expression such that it must be less than [MATH]\frac{12}{25}[/MATH].

As a matter of fact, we could proceed to find the maximum for that target expression without using $a+b+c+d=4$, but we most likely would end up with getting to prove it less than or equal to a value much greater than [MATH]\frac{12}{25}[/MATH], so that won't help.

Question 2: Given a is a positive real number. What other equivalent you could think of to represent [MATH]\frac{a}{a^3+8}[/MATH]?

A. [MATH]\frac{a}{a^3+8}=\frac{1}{a^2}+\frac{a}{8}[/MATH].
B. [MATH]\frac{a}{a^3+8}=\frac{1}{a^2+\frac{8}{a}}[/MATH].
C. [MATH]\frac{a}{a^3+8}=\frac{a}{2(a^3+4)}+\frac{a}{2(a^3+4)}[/MATH].

Note that [MATH]\frac{a}{a^3+8}=\frac{1}{\frac{a^3+8}{a}}=\frac{1}{\frac{a^3}{a}+\frac{8}{a}}=\frac{1}{a^2+\frac{8}{a}}[/MATH], therefore B is the correct answer.

Question 3: If we want to mazimize [MATH]\frac{a}{a^3+8}[/MATH] and [MATH]\frac{a}{a^3+8}=\frac{1}{a^2+\frac{8}{a}}[/MATH], what would you do to the expression [MATH]a^2+\frac{8}{a}[/MATH] in order to maximize [MATH]\frac{a}{a^3+8}[/MATH]?

A. Minimize [MATH]a^2+\frac{8}{a}[/MATH].
B. Minimize [MATH]\frac{1}{a^2+\frac{8}{a}}[/MATH].
C. Maximize [MATH]a^2+\frac{8}{a}[/MATH].
D. Maximize [MATH]\frac{1}{a^2+\frac{8}{a}}[/MATH].

We know [MATH]\frac{a}{a^3+8}=\frac{1}{a^2+\frac{8}{a}}[/MATH], so in order to maximize [MATH]\frac{a}{a^3+8}[/MATH], i.e. [MATH]\frac{1}{a^2+\frac{8}{a}}[/MATH] we have got to minimize [MATH]a^2+\frac{8}{a}[/MATH]. Therefore the answer for this question is A.

Question 4: Determine which of the following inequality holds for positive $a$?

[MATH]a^2+\frac{8}{a}\ge \frac{4\sqrt{2}}{\sqrt{a}}[/MATH]
[MATH]a^2+\frac{8}{a}\ge \frac{2\sqrt{2}}{\sqrt{a}}[/MATH]
[MATH]a^2+\frac{8}{a}\ge 4\sqrt{2a}[/MATH]
[MATH]a^2+\frac{8}{a}\ge 6\sqrt[3]{2}[/MATH]
[MATH]a^2+\frac{8}{a}\ge \frac{9}{a^{\frac{2}{3}}}[/MATH]
[MATH]a^2+\frac{8}{a}\ge 4\left(\frac{8^3}{3^3a}\right)^{\frac{1}{4}}[/MATH]

By using AM-GM inequality, we see that:

[MATH]\begin{align*}a^2+\frac{8}{a}&\ge 2\sqrt{a^2\cdot \frac{8}{a}}\\&\ge 2\cdot 2\sqrt{2a}\\&\ge 4\sqrt{2a}\end{align*}[/MATH]

Also, if we rewrite [MATH]a^2+\frac{8}{a}[/MATH] as [MATH]a^2+\frac{4}{a}+\frac{4}{a}[/MATH], we would get:

[MATH]\begin{align*}a^2+\frac{8}{a}=a^2+\frac{4}{a}+\frac{4}{a}&\ge 3\sqrt[3]{a^2\cdot \frac{4}{a}\cdot \frac{4}{a}}\\&\ge 3\sqrt[3]{16}\\&\ge 3\sqrt{2^3\cdot 2}\\&\ge 3\cdot 2 \sqrt[3]{2}\\&\ge 6\sqrt[3]{2}\end{align*}[/MATH]

Next, if we replace [MATH]a^2+\frac{8}{a}[/MATH] by [MATH]a^2+\frac{8}{3a}+\frac{8}{3a}+\frac{8}{3a}[/MATH], this yields

[MATH]\begin{align*}a^2+\frac{8}{a}=a^2+\frac{8}{3a}+\frac{8}{3a}+\frac{8}{3a}&\ge 4\sqrt[4]{a^2\cdot \frac{8}{3a}\cdot \frac{8}{3a}\cdot \frac{8}{3a}}\\&\ge 4\sqrt[4]{\frac{8^3}{3^3a}}\end{align*}[/MATH]

Now, if we rewrite [MATH]a^2+\frac{8}{a}[/MATH] as [MATH]a^2+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}[/MATH], we would end up with:

[MATH]\begin{align*}a^2+\frac{8}{a}&=a^2+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}\\&\ge 9\sqrt[9]{a^2\cdot \frac{1}{a}\cdot \frac{1}{a}\cdot \frac{1}{a}\cdot \frac{1}{a}\cdot \frac{1}{a}\cdot \frac{1}{a}\cdot \frac{1}{a}}\\&\ge 9\sqrt[9]{\frac{1}{a^6}}\\&\ge \frac{9}{a^{\frac{6}{9}}}\\&\ge \frac{9}{a^{\frac{2}{3}}}\end{align*}[/MATH]

Therefore, one should have checked the last four options as the correct answers.

Question 5: What type of inequality formula would help you to determine the correct answer(s) for question 4?

A. AM-GM inequality.
B. Cauchy–Schwarz inequality
C. Power Mean inequality.
D. Jensen's inequality.

We used merely AM-G inequality in our working to look for the correct answers in question 4. Therefore the answer is A.

Question 6: It's well-known from the Cauchy–Schwarz inequality that for all real numbers $a,\,b$ and $c$, we have $a+b+c\le\sqrt{3}\sqrt{a^2+b^2+c^2}$.

Do you think $a+b+c+d\le \sqrt{4}\sqrt{a^2+b^2+c^2+d^2}$ is also true for all real numbers $a,\,b,\,c$ and $d$?

A. Yes.
B. No.

According to the Cauchy–Schwarz inequality, if we have four real numbers $a,\,b,\,c$ and $d$, then we have:

\begin{align*}a+b+c+d=1(a)+1(b)+1(c)+1(d)&\le \sqrt{1^2+1^2+1^2+1^2}\sqrt{a^2+b^2+c^2+d^2}\\&\le \sqrt{4}\sqrt{a^2+b^2+c^2+d^2}\end{align*}

Therefore, the answer for this question is a Yes.

Question 7: Which of the answer from question 4 would you have picked to prove the inequality as stated in question 1?

Hint: We want to use the equality $a+b+c+d=4$ in our solution.

A. [MATH]a^2+\frac{8}{a}\ge 4\sqrt{2a}[/MATH]

B. [MATH]a^2+\frac{8}{a}\ge 6\sqrt[3]{2}[/MATH]

C. [MATH]a^2+\frac{8}{a}\ge \frac{9}{a^{\frac{2}{3}}}[/MATH]

D. [MATH]a^2+\frac{8}{a}\ge 4\left(\frac{8^3}{3^3a}\right)^{\frac{1}{4}}[/MATH]

We know we must make use of the equality $a+b+c+d=4$ if we want to find the maximum of[MATH]\sum_{cyclic} \frac{a}{a^3+8}[/MATH], therefore, we need to rewrite it in such a way that after applying for the Cauchy–Schwarz inequality, we would wind up getting with the final expression that has only the sum of $a+b+c+d$ raised to some powers.

It requires us to do some investigation before realizing D is the answer, since:

[MATH]\sum_{cyclic} \frac{a}{a^3+8}[/MATH]

[MATH]=\sum_{cyclic}\frac{1}{a^2+\frac{8}{a}}[/MATH]

[MATH]=\frac{1}{a^2+\frac{8}{a}}+\frac{1}{b^2+\frac{8}{b}}+\frac{1}{c^2+\frac{8}{c}}+\frac{1}{d^2+\frac{8}{d}}[/MATH]

[MATH]\le \frac{1}{4\left(\frac{8^3}{3^3a}\right)^{\frac{1}{4}}}+\frac{1}{4\left(\frac{8^3}{3^3b}\right)^{\frac{1}{4}}}+\frac{1}{4\left(\frac{8^3}{3^3c}\right)^{\frac{1}{4}}}+\frac{1}{4\left(\frac{8^3}{3^3d}\right)^{\frac{1}{4}}}[/MATH]

[MATH]\le \frac{1}{4}\left(\frac{3}{8}\right)^{\frac{3}{4}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}+c^{\frac{1}{4}}+d^{\frac{1}{4}}\right)[/MATH]

Therefore, the answer for this question 7 is D.

Question 8: Do you think you can now prove [MATH]\sum_{cyclic} \frac{a}{a^3+8}\le\frac{4}{9}[/MATH]., given $a,\,b,\,c$ and $d$ are all positive real numbers such that $a+b+c+d=4$?

A. Yes.
B. No.

When you have went through the previous questions and thought reasonably and logically for each question, then you would be able to prove the required inequality without trouble.

Let's take up where we left off from the previous question:

[MATH]\sum_{cyclic} \frac{a}{a^3+8}[/MATH]

[MATH]\le \frac{1}{4}\left(\frac{3}{8}\right)^{\frac{3}{4}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}+c^{\frac{1}{4}}+d^{\frac{1}{4}}\right)[/MATH]

[MATH]\le \frac{1}{4}\left(\frac{3}{8}\right)^{\frac{3}{4}}\sqrt{1+1+1+1}\sqrt{a^{\frac{1}{2}}+b^{\frac{1}{2}}+c^{\frac{1}{2}}+d^{\frac{1}{2}}}[/MATH] (by the Cauchy–Schwarz inequality)

[MATH]\le \frac{1}{4}\left(\frac{3}{8}\right)^{\frac{3}{4}}\sqrt{4}\sqrt{\sqrt{1+1+1+1}\sqrt{a+b+c+d}}[/MATH] (again by the Cauchy–Schwarz inequality)

[MATH]\le \frac{1}{4}\left(\frac{3}{8}\right)^{\frac{3}{4}}\cdot 2 \sqrt{\sqrt{4}\sqrt{4}}[/MATH]

[MATH]\le \frac{1}{4}\left(\frac{3}{8}\right)^{\frac{3}{4}}\cdot 4[/MATH]

[MATH]\le \left(\frac{3}{8}\right)^{\frac{3}{4}}[/MATH]

Since $3^3(25^4)=10,546,875 \lt (12^4)(8^3)=10,016,832$, we have $\left(\dfrac{3}{8}\right)^{\frac{3}{4}}\lt \dfrac{12}{25}$ and therefore we have proved that

[MATH]\sum_{cyclic} \frac{a}{a^3+8}\le \left(\frac{3}{8}\right)^{\frac{3}{4}}\lt \dfrac{12}{25}[/MATH]