Explore Fraction Concepts with Examples and Questions

Write each fraction in terms of the LCD.
$$\frac{3}{a{b}^2};\frac{7}{ab}$$

Proving an inequality via the Cauchy-Schwarz Inequality
I apolgize for contributing yet another question asking about an application of CS. Here it is:
Suppose $p_1,\dots <!--\; \dots \; -->,p_n$ and $a_1,...,a_n$ are real numbers such that $p_i\ge <!--\; \ge \; -->0$, $a_i\ge <!--\; \ge \; -->0$ for all $i$, and $p_1+\cdots <!--\; \cdots \; -->+p_n=1$. Then
$(p_1{a}_1+\cdots <!--\; \cdots \; -->+p_n{a}_n)(\frac{p_1}{a_1}+\cdots <!--\; \cdots \; -->+\frac{p_n}{a_n})\ge <!--\; \ge \; -->1$
The author of my textbook gives the following proof: Apply Cauchy's inequality to the sequences $\sqrt{p_1{a}_1}\dots <!--\; \dots \; -->\sqrt{p_n}{a}_n$ and $\sqrt{\frac{p_1}{a_1}}\dots <!--\; \dots \; -->\sqrt{\frac{p_n}{a_n}}$. (Thats it)
In trying to fill in the blanks I obtained the following
$\sqrt{p_1{a}_1}+\cdots <!--\; \cdots \; -->+\sqrt{p_n{a}_n}\le <!--\; \le \; -->\sqrt{p_1+\cdots <!--\; \cdots \; -->+p_n}\sqrt{a_1+...+a_n}=\sqrt{a_1+...+a_n}$
and
$\sqrt{\frac{p_1}{a_1}}+\cdots <!--\; \cdots \; -->+\sqrt{\frac{p_n}{a_n}}\le <!--\; \le \; -->\sqrt{p_1+\cdots <!--\; \cdots \; -->+p_n}\sqrt{\frac{1}{a_1}+...+\frac{1}{a_n}}=\sqrt{\frac{1}{a_1}+...+\frac{1}{a_n}}$
I'm not entirely sure where to go from here. Perhaps I have misunderstood what he meant by "apply cauchy's inequality to the sequences...". Another idea I had was to note that
$(p_1{a}_1+\cdots <!--\; \cdots \; -->+p_n{a}_n)\le <!--\; \le \; -->M_a(p_1+\cdots <!--\; \cdots \; -->+p_n)$
where $M_a$ is the largest $a_i$. And, that
$(\frac{p_1}{a_1}+\cdots <!--\; \cdots \; -->+\frac{p_n}{a_n})\le <!--\; \le \; -->\frac{1}{m_a}(p_1+\cdots <!--\; \cdots \; -->+p_n)$
where $m_a$ is the smallest $a_i$. Therefore, since $\frac{M_a}{m_a}\ge <!--\; \ge \; -->1$ the inequality follows. I am not very confident in the correctness of this method though and would like to understand how to prove the inequality via CS as my book suggests.

I'm reviewing my arithmetic and right now I'm at fractions, I'm just having a little bit of problem "visualizing" why cancellation works in multiplying fractions, I know how it works, it's just the why. Take for example.
$$\frac{2}{3}\cdot <!--\; \cdot \; -->\frac{3}{4}=\frac{1}{3}\cdot <!--\; \cdot \; -->\frac{3}{2}=\frac{1\cdot <!--\; \cdot \; -->1}{1\cdot <!--\; \cdot \; -->2}=\frac{1}{2}.$$
Through cancellation we know this ends in 1/2 because the common factor of the numerator "2" and denomaninator "4" is 2, so we divide both by "2" and end with new numbers in place, we also know that the common factor of numerator "3" and denominator "3" is "3" which equals "1" in both places, so it makes the work of "reducing to lowest terms" non-existent since this method is already a shortcut to that.
But I just can't seem to visualize how it works, can you guys give me a visual model to help me make sense of this? thank you.

Exponent Simplification
I have this expression where u and k are arbitrary constants:
$u^{{\left(\frac{1}{k}\right)}^{{\log }_k<!--\; \; -->({\log }_k<!--\; \; -->(u))}}$
I'm trying to clean up or simplify this expression... how can i go just making this cleaner? I'm forgetting log and exponent rules. Thanks!

Problem proving inequality $\frac{4^n}{n+1}\le <!--\; \le \; -->\frac{(2n)!}{(n!{)}^2}$
I skip the base case $n=0$ because it's obvious.
I know that this is very equivalent to: Prove by induction: $\frac{2^{2n}}{n+1}<\frac{(2n)!}{(n!{)}^2},n>1$
But I try to learn some tricks so maybe you can help me?
Assumption: $\frac{4^n}{n+1}\le <!--\; \le \; -->\frac{(2n)!}{(n!{)}^2}⟺<!--\; ⟺\; -->4^n\le <!--\; \le \; -->\frac{(n+1)(2n)!}{(n!{)}^2}$ for some n.
Step: $n\to <!--\; \to \; -->n+1$
$4^{n+1}=\frac{(n+2)(2n+2)!}{((n+1)!{)}^2}$
I begin with manipulating the LHS:
$4^{n+1}=4\cdot <!--\; \cdot \; -->4^n\le <!--\; \le \; -->???$
$???\le <!--\; \le \; -->\frac{(2n)!(2n+2)(2n+1)(n+1)}{(n!(n+1))}$
$=\frac{(n+1)(2n+2)!}{((n+1)!{)}^2}$
Note that I want to work my way from both sides to the mid where I get a very easy to check inequality by assumption.
Can you help me?

Do there exist positive integers $a,b,c$ such that $1/a+1/b+1/c=47/48$?
So far the only conclusion i have drawn is that $a+b+c$ is greater than 48. However, I cannot make anymore connections that would help me solve the problem. Am I just supposed to "brute force" this by bashing the algebra, or is there some logic I am missing?

Divide and simplify.
$$\frac{x^2+4}{8(x+3)}÷<!--\; ÷\; -->\frac{x+2}{8(x^2-<!--\; -\; -->9)}$$

Find the minimum value of an expression with three variables
How can I find the minimum of the following expression:
$(\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x})(\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy})$
($x,y,z$ are non-zero real numbers)
The expression can be simplified to $\frac{((xy{)}^2+(zx{)}^2+(yz{)}^2)(x^2+y^2+z^2)}{(xyz{)}^2}$, but I am not sure that it will help anyhow.

Why can't I separate the fraction inside a fraction to the outside
I know when I have something of this form
$$a\frac{1}{\mathrm{\alpha <!--\; \alpha \; -->}}\frac{s+\frac{1}{T}}{s+\frac{1}{\mathrm{\alpha <!--\; \alpha \; -->}T}}$$
I multiply out the fractional part, α with the bottom to get
$$a\frac{s+\frac{1}{T}}{\mathrm{\alpha <!--\; \alpha \; -->}s+\frac{1}{T}}$$
But, if I instead have
$$a\frac{\frac{1}{\mathrm{\alpha <!--\; \alpha \; -->}}(s+\frac{1}{T})}{s+\frac{1}{\mathrm{\alpha <!--\; \alpha \; -->}T}}$$
with the fractional part
$$\frac{1}{\mathrm{\alpha <!--\; \alpha \; -->}}$$
in the numerator instead.
I know I can't take the fractional part out of the equation in this second instance. But, I don't know why that is. I know this is a fairly basic rule, but I haven't done algebra in quite a while. So, I wanted to ask for clarification.

Simplifying an expression: Stuck with it
I have to prove that the expression
$$\frac{\mathrm{\omega <!--\; \omega \; -->}C-<!--\; -\; -->\frac{1}{\mathrm{\omega <!--\; \omega \; -->}L}}{\mathrm{\omega <!--\; \omega \; -->}C-<!--\; -\; -->\frac{1}{\mathrm{\omega <!--\; \omega \; -->}L}+\mathrm{\omega <!--\; \omega \; -->}L-<!--\; -\; -->\frac{1}{\mathrm{\omega <!--\; \omega \; -->}C}}$$
is equal to
$$\frac{1}{3-<!--\; -\; -->((\frac{{\mathrm{\omega <!--\; \omega \; -->}}_r}{\mathrm{\omega <!--\; \omega \; -->}}{)}^2+(\frac{\mathrm{\omega <!--\; \omega \; -->}}{{\mathrm{\omega <!--\; \omega \; -->}}_r}{)}^2)}$$
where ${\mathrm{\omega <!--\; \omega \; -->}}_r=\frac{1}{\sqrt{LC}}$.

Fraction rules A/B/C vs B/C/A
I have all my fraction rules written down, but these two confuse me...
$$\frac{A}{\frac{B}{C}}\equiv <!--\; \equiv \; -->\frac{AC}{B}$$
and
$$\frac{\frac{B}{C}}{A}\equiv <!--\; \equiv \; -->\frac{B}{CA}$$
These do not give the same answers, and I am wondering what the rules are for using them? Are
$$A>B>C$$
and
$$B>C>A$$
the two conditions?

Approximation for the infinite counting decimal
Somewhere between the real number range, there exists a decimal that 'counts' natural numbers infinitely on its digits as:
$\mathrm{0.123456789101112131415161718192021.........}$
It goes on 'counting' forever.
It is an irrational number, so
I'd like to find a close approximation of it (for about 30+ digits for example) like the PI approximation $(22/7)$
One last question for fun:
Is that number has been given a name before?

Sum of Unit Fractions Equal to 1 Has Finitely Many Solutions
My book on abstract algebra had a small puzzle in it I was interested in. It asks the reader to prove that
$\underset{j=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{1}{x_j}=1$ only has finitely many solutions where $x_j\in <!--\; \in \; -->\mathbb{N}$
My first attempt to proof it was with induction. It's easy to see that for $n=1$ it only has $1$ solution. For the induction step I assume that in the case $n=k+1$ we have infinitely many solutions, so it follows that $\underset{j=1}{\overset{k}{\sum <!--\; \sum \; -->}}\frac{1}{x_j}=\frac{x_k-<!--\; -\; -->1}{x_k}$ which implies that $\underset{j=1}{\overset{k}{\sum <!--\; \sum \; -->}}\frac{1}{x_j}=\frac{1}{x_k}$ (assuming $x_k\ne <!--\; \ne \; -->1$).
This is pretty much as far as I got. I'd really love to have a hint rather than a full solution if possible.

Inequality $\frac{x_1^2}{x_1^2+x_2{x}_3}+\frac{x_2^2}{x_2^2+x_3{x}_4}+\cdots <!--\; \cdots \; -->+\frac{x_{n-<!--\; -\; -->1}^2}{x_{n-<!--\; -\; -->1}^2+x_n{x}_1}+\frac{x_n^2}{x_n^2+x_1{x}_2}\le <!--\; \le \; -->n-<!--\; -\; -->1$
Show that for all $n\ge <!--\; \ge \; -->2$
$\frac{x_1^2}{x_1^2+x_2{x}_3}+\frac{x_2^2}{x_2^2+x_3{x}_4}+\cdots <!--\; \cdots \; -->+\frac{x_{n-<!--\; -\; -->1}^2}{x_{n-<!--\; -\; -->1}^2+x_n{x}_1}+\frac{x_n^2}{x_n^2+x_1{x}_2}\le <!--\; \le \; -->n-<!--\; -\; -->1$
where $x_i$ are real positive numbers
I was going to use $\frac{x_1^2}{x_1^2+x_2{x}_3}=\frac{1}{1+\frac{x_2{x}_3}{x_1^2}}\le <!--\; \le \; -->\frac{x_1}{2}\cdot <!--\; \cdot \; -->\frac{1}{\sqrt{x_2{x}_3}}\le <!--\; \le \; -->\frac{x_1}{4}(\frac{1}{x_2}+\frac{1}{x_3})$
$\frac{x_1^2}{x_1^2+x_2{x}_3}+\frac{x_2^2}{x_2^2+x_3{x}_4}+...+\frac{x_{n-<!--\; -\; -->2}^2}{x_{n-<!--\; -\; -->2}^2+x_{n-<!--\; -\; -->1}{x}_n}+\frac{x_{n-<!--\; -\; -->1}^2}{x_{n-<!--\; -\; -->1}^2+x_n{x}_1}+\frac{x_n^2}{x_n^2+x_1{x}_2}$
$=\frac{1}{1+\frac{x_2{x}_3}{x_1^2}}+\frac{1}{1+\frac{x_3{x}_4}{x_2^2}}+...+\frac{1}{1+\frac{x_{n-<!--\; -\; -->1}{x}_n}{x_{n-<!--\; -\; -->2}^2}}+\frac{1}{1+\frac{x_n{x}_1}{x_{n-<!--\; -\; -->1}^2}}+\frac{1}{1+\frac{x_1{x}_2}{x_n^2}}$
$\le <!--\; \le \; -->\frac{x_1}{4}(\frac{1}{x_2}+\frac{1}{x_3})+\frac{x_2}{4}(\frac{1}{x_3}+\frac{1}{x_4})+...+\frac{x_{n-<!--\; -\; -->2}}{4}(\frac{1}{x_{n-<!--\; -\; -->1}}+\frac{1}{x_n})+\frac{x_{n-<!--\; -\; -->1}}{4}(\frac{1}{x_n}+\frac{1}{x_1})+\frac{x_n}{4}(\frac{1}{x_1}+\frac{1}{x_2})$
$=\frac{1}{4}((\frac{x_1}{x_2}+\frac{x_1}{x_3})+(\frac{x_2}{x_3}+\frac{x_2}{x_4})+(\frac{x_3}{x_4}+\frac{x_3}{x_5})+...+(\frac{x_{n-<!--\; -\; -->2}}{x_{n-<!--\; -\; -->1}}+\frac{x_{n-<!--\; -\; -->2}}{x_n})+(\frac{x_{n-<!--\; -\; -->1}}{x_n}+\frac{x_{n-<!--\; -\; -->1}}{x_1})+(\frac{x_n}{x_1}+\frac{x_n}{x_2}))$
$=\frac{1}{4}(\left(\frac{x_1+x_2}{x_3}\right)+\left(\frac{x_2+x_3}{x_4}\right)+\left(\frac{x_3+x_4}{x_5}\right)+...+\left(\frac{x_{n-<!--\; -\; -->3}+x_{n-<!--\; -\; -->2}}{x_{n-<!--\; -\; -->1}}\right)+\left(\frac{x_{n-<!--\; -\; -->1}+x_{n-<!--\; -\; -->2}}{x_n}\right)+\left(\frac{x_1+x_n}{x_2}\right)+\left(\frac{x_{n-<!--\; -\; -->1}+x_n}{x_1}\right))$
....I thought about using Cauchy's inequality, but that would only increase the problem

Prove that $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\left(\frac{x_i}{x_{i+1}}\right)}^4\ge <!--\; \ge \; -->\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\frac{x_i}{x_{i+4}}$
How to prove this inequality?
${\left(\frac{x_1}{x_2}\right)}^4+{\left(\frac{x_2}{x_3}\right)}^4+\cdots <!--\; \cdots \; -->+{\left(\frac{x_n}{x_1}\right)}^4\ge <!--\; \ge \; -->\frac{x_1}{x_5}+\frac{x_2}{x_6}+\cdots <!--\; \cdots \; -->+\frac{x_{n-<!--\; -\; -->3}}{x_1}+\frac{x_{n-<!--\; -\; -->2}}{x_2}+\frac{x_{n-<!--\; -\; -->1}}{x_3}+\frac{x_n}{x_4},$
where $x_1,x_2,\dots <!--\; \dots \; -->,x_n>0.$

Have I made a new limit that tends to $e$?
I have made two formulae. One tending to $e$ and one tending to $3$:
$e=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\left(\frac{2\times <!--\; \times \; -->{10}^n-<!--\; -\; -->1}{2\times <!--\; \times \; -->{10}^n-<!--\; -\; -->3}\right)}^{{10}^n-<!--\; -\; -->1}$
$3=\underset{n\to <!--\; \to \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->})}{lim}{\left(\frac{2\times <!--\; \times \; -->{10}^n-<!--\; -\; -->1}{2\times <!--\; \times \; -->{10}^n-<!--\; -\; -->3}\right)}^{{10}^n-<!--\; -\; -->1}$
Is there a way of proving that these formulae are indeed true? Thanks.

Moving decimal on multiplying or dividing by power of 10
How does "moving decimal" works on multiplying or dividing by any number which is power of 10? For example: 1.23×100 = 123 in this example only value is changed while digits remains same

To find the Modulus of a complex number
I was given the following expression ,
$|\frac{(3+i)(2-<!--\; -\; -->i)}{(1+i)}|$
and was asked to find its value
This is how I proceeded ,
On solving the numerator , the given expression transforms to $|\frac{7-<!--\; -\; -->i}{1+i}|$ Then I took the conjugate of the denominator and finally got the expression
$\frac{8-<!--\; -\; -->8i}{2}$
$=4|(1-<!--\; -\; -->i)|$
Now according to me it’s modulus should be $4\sqrt{2}$ however the correct answer is $5$ . Could you please correct me where I am mistaken ? And please suggest a method to solve this. Thank you .

Given the positive numbers $a,b,c$. Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le <!--\; \le \; -->\frac{3}{2}$
By Cauchy Schwarz, we have that
$(a^2+1)(1+3)\ge <!--\; \ge \; -->{(a+\sqrt{3})}^2\to <!--\; \to \; -->\frac{a}{\sqrt{a^2+1}}\le <!--\; \le \; -->\frac{2a}{a+\sqrt{3}}$
I need a new method

Prove $\frac{1}{2}+\frac{1}{2(u+1{)}^2}-<!--\; -\; -->\frac{1}{\sqrt{1+2u}}\ge <!--\; \ge \; -->0$ for $u\ge <!--\; \ge \; -->0$
This inequality provides a tight lower bound to $\sqrt{1+2u}$ for $u\ge <!--\; \ge \; -->0$ without a radical. I was trying to solve it by squaring the radical and cross-multiplying and repeated differentiation of the resulting expression, I wonder if there is a quicker solution. Thanks.