Explore Fraction Concepts with Examples and Questions

Four positive numerator-1 fractions summing to 3/7
This year's seventh-grade olympiad sponsored by Tel Aviv University, round two, held a couple of days ago, had this (translated by yours truly) as its third question:
Find four distinct natural numbers, $a$, $b$, $c$, and $d$, such that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{3}{7}$
My solution:
$\begin{array}{rl}\frac{3}{7}& =\frac{1}{7}+.\stackrel{¯<!--\; ¯\; -->}{285714}\\ & =\frac{1}{7}+\frac{1}{5}+.0\stackrel{¯<!--\; ¯\; -->}{857142}\\ & =\frac{1}{7}+\frac{1}{5}+\frac{6}{70}\\ & =\frac{1}{7}+\frac{1}{5}+\frac{1}{70}+\frac{1}{14}\end{array}$
Tweaking that a bit led me to
$\frac{1}{7}+\frac{1}{4}+\frac{1}{32}+\frac{1}{224}$
Question: Are there more solutions? Especially: Are there infinitely many?

Which of the following pairs of numbers is between -3 and -2?
a) $-<!--\; -\; -->2\frac{1}{2}$ , $-<!--\; -\; -->2\frac{1}{3}$
b) -2, $-<!--\; -\; -->3\frac{1}{3}$
c) $-<!--\; -\; -->3.14,-<!--\; -\; -->2.184$
d) $-<!--\; -\; -->2\frac{1}{2},-<!--\; -\; -->3$

Calculate ${\tan }^2<!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{5}+{\tan }^2<!--\; \; -->\frac{2\mathrm{\pi <!--\; \pi \; -->}}{5}$ without a calculator
The question is to find the exact value of:
${\tan }^2<!--\; \; -->\left(\frac{\mathrm{\pi <!--\; \pi \; -->}}{5}\right)+{\tan }^2<!--\; \; -->\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{5}\right)$
without using a calculator.
I know that it is possible to find the exact values of $\tan <!--\; \; -->\left(\frac{\mathrm{\pi <!--\; \pi \; -->}}{5}\right)$ and $\tan <!--\; \; -->\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{5}\right)$ to find that the answer is $10$. However, I want to know whether there is a faster way that does not involve calculating those values.
So far, I have this: let $a=\tan <!--\; \; -->\left(\frac{\mathrm{\pi <!--\; \pi \; -->}}{5}\right)$ and $b=\tan <!--\; \; -->\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{5}\right)$; then, $b=\frac{2a}{1-<!--\; -\; -->a^2}$ and $a=-<!--\; -\; -->\frac{2b}{1-<!--\; -\; -->b^2}$, so multiplying the two and simplifying gives:
$a^2+b^2={\left(ab\right)}^2+5$
Any ideas? Thanks!

Show that if $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$, then $\frac{1}{3+a}+\frac{1}{3+c}+\frac{1}{3+c}\le <!--\; \le \; -->\frac{3}{4}$
The corresponding problem replacing the $3$s with $2$ is shown here:
How to prove $\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}\le <!--\; \le \; -->1$?
The proof is not staggeringly difficult: The idea is to show (the tough part) that if $\frac{1}{2+a}+\frac{1}{2+c}+\frac{1}{2+c}=1$ then $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge <!--\; \ge \; -->a+b+c$. The result then easily follows.
A pair of observations, both under the constraint of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$:
If $k\ge <!--\; \ge \; -->2$ then $\frac{1}{k+a}+\frac{1}{k+c}+\frac{1}{k+c}\le <!--\; \le \; -->\frac{3}{k+1}$. This is proven for $k=2$ but I don't see how for $k>2$
If $0<k<2$ then there are positive $(a,b,c)$ satisfying the constraint such that $\frac{1}{k+a}+\frac{1}{k+c}+\frac{1}{k+c}>\frac{3}{k+1}$
So one would hope that the $k=3$ case, lying farther within the "valid" region, might be easier than $k=2$ but I have not been able to prove it.
Note that it is not always true, under our constraint, that $\frac{1}{2+a}+\frac{1}{2+c}+\frac{1}{2+c}\ge <!--\; \ge \; -->\frac{1}{3+a}+\frac{1}{3+c}+\frac{1}{3+c}+\frac{1}{4}$, which if true would prove the $k=3$ case immediately.

What's wrong with my workings? Adding fractions question.
Please could someone explain what is wrong with my workings?
$(\frac{1}{9}+\frac{1}{2})+(\frac{1}{3}-<!--\; -\; -->\frac{1}{2})=?$
My workings are:
$\left(18(\frac{1}{9}+\frac{1}{2})\right)+\left(6(\frac{1}{3}-<!--\; -\; -->\frac{1}{2})\right)=(2+6)+(2-<!--\; -\; -->3)=8-<!--\; -\; -->1=7$
WHERE DID I GO WRONG ?

minimum value of ${\displaystyle f(x)=\frac{(1+x{)}^{0.8}}{1+x^{0.8}}}$ without derivative
minimum value of ${\displaystyle f(x)=\frac{(1+x{)}^{0.8}}{1+x^{0.8}},x\ge <!--\; \ge \; -->0}$ without derivative
Binomial expansion of ${\displaystyle (1+x{)}^{0.8}=1+0.8x-<!--\; -\; -->\frac{(0.8\cdot <!--\; \cdot \; -->(0.8-<!--\; -\; -->1)x^2)}{2}+\cdots <!--\; \cdots \; -->}$
but from above does not get anything
could some help me with this

Problem: Log function with common base equation
ln(x-2)-ln(x+9)=ln(x-1)-ln(x+14)
I dropped the ln's and tried solving by using the quotient property but I think im doing the fractions wrong.
can someone please explain how to get the answer? (solving for x)
Thank you

How is this equation true?
I was looking through my notes when I stumbled upon this eqation:
$\sqrt{n^2+n}-<!--\; -\; -->n=\frac{(n^2+n)-<!--\; -\; -->n^2}{\sqrt{n^2+n}+n}$
I was trying to make sense of it, but I didnt succeed.
Can you give me a hint or tell me how this tranformation is called ?

Prove inequality $\sum <!--\; \sum \; -->\frac{a^3-<!--\; -\; -->b^3}{{\left(a-<!--\; -\; -->b\right)}^3}\ge <!--\; \ge \; -->\frac{9}{4}$
Given $a,b,c$ are positive number. Prove that
$\frac{a^3-<!--\; -\; -->b^3}{{(a-<!--\; -\; -->b)}^3}+\frac{b^3-<!--\; -\; -->c^3}{{(b-<!--\; -\; -->c)}^3}+\frac{c^3-<!--\; -\; -->a^3}{{(c-<!--\; -\; -->a)}^3}\ge <!--\; \ge \; -->\frac{9}{4}$
$\iff <!--\; \iff \; -->\sum <!--\; \sum \; -->\frac{3(a+b{)}^2+(a-<!--\; -\; -->b{)}^2}{(a-<!--\; -\; -->b{)}^2}\ge <!--\; \ge \; -->9$
$\iff <!--\; \iff \; -->\frac{(a+b{)}^2}{(a-<!--\; -\; -->b{)}^2}+\frac{(b+c{)}^2}{(b-<!--\; -\; -->c{)}^2}+\frac{(c+a{)}^2}{(c-<!--\; -\; -->a{)}^2}\ge <!--\; \ge \; -->2$
Which
$\frac{a+b}{a-<!--\; -\; -->b}.\frac{b+c}{b-<!--\; -\; -->c}+\frac{b+c}{b-<!--\; -\; -->c}.\frac{c+a}{c-<!--\; -\; -->a}+\frac{c+a}{c-<!--\; -\; -->a}.\frac{a+b}{a-<!--\; -\; -->b}=-<!--\; -\; -->1$
Use $x^2+y^2+z^2\ge <!--\; \ge \; -->-<!--\; -\; -->2(xy+yz+zx)$ then inequality right
I don't know why use $x^2+y^2+z^2\ge <!--\; \ge \; -->-<!--\; -\; -->2(xy+yz+zx)$ then inequality right?
P/s: Sorry i knowed, let $x=\frac{a+b}{a-<!--\; -\; -->b}$
We have $(x+1)(y+1)(z+1)=(x-<!--\; -\; -->1)(y-<!--\; -\; -->1)(z-<!--\; -\; -->1)=>xy+yz+zx=-<!--\; -\; -->1$

why does multiplying a fraction e.g. enumerator/divisor with divisor/enumerator give 1?
I just found out that if you want to get 1 with the fraction:
$\frac{5}{2}$
Then you multiply it with:
$\frac{2}{5}$
Does anyone have a good way to think about this?

Put fraction in "arctan-friendly" form
I would like to put $\int <!--\; \int \; -->\frac{1}{(2x^2+x+1)}dx$ into something like $\int <!--\; \int \; -->\frac{1}{(u^2+1)}dx$. What is the quickest way to proceed? I know that previous fraction can be rewritten as $2t^2+t+1=\frac{7}{8}({\left(\frac{4t+1}{\sqrt{7}}\right)}^2+1)$, but I don't have any explaination from where this comes from.
Finally, the integral yields
${\int <!--\; \int \; -->}_b^a\frac{7}{8}({\left(\frac{4t+1}{\sqrt{7}}\right)}^2+1)dt=\frac{2}{\sqrt{7}}{[\arctan <!--\; \; -->\left(\frac{4t+1}{\sqrt{7}}\right)]}_b^a$

Proving inequality using Lagrange multipliers.
I started learing about Lagrange Multipliers and I got the following question:
Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$, for each $a,b,c>0$
I'm not sure how to use Lagrange multipliers for it... any ideas?

What does $2^{2.5}$ mean ?
We know that $2^2$ means $2\cdot <!--\; \cdot \; -->2$, $2^3$ means $2\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->2$, and $2^5$ means $2\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->2$ and so on and so forth. But what does $2^{2.5}$ mean? I don't mean to say "How to find it out". I mean, what is the idea behind it? Is it something like "If $S=V\cdot <!--\; \cdot \; -->T$, then $T=\frac{S}{V}$ and $T$ gets smaller as $V$ gets bigger, and theoretically $V$ can get so big that $T$ might become $0$. (that is, $\underset{V\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{S}{V}=0,(T=\frac{S}{V}))$" but can't be shown physically?

How do I express 0.999(9) as a fraction?
I'm noob in math. If 0.333(3) is 1/3, 0.666(6) is 2/3, then 0.999(9) is what?
If 3/3 and 0.999(9) is the same, then how can I express one of them without expressing the other?

Simplifying $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\sqrt[n]{2}-<!--\; -\; -->1}{\sqrt[n]{8}-<!--\; -\; -->1}$
I was attempting to go through a worked solution from my lecturer but I can't really understand this step:
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\sqrt[n]{2}-<!--\; -\; -->1}{\sqrt[n]{8}-<!--\; -\; -->1}=\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{1}{\sqrt[n]{4}+\sqrt[n]{2}+1}$
What was actually even done here? And why?
Any help is appreciated.

Question trying to physically understand Algorithm run time as proportion
In CLRS, there's a question in the first chapter that says for certain functions $runtime=f(n)$, to calculate the max number of $n$ given the run time. The math works out simply. For example, for $f(n)=\sqrt{n}$ microseconds, given a 1 second run time there is ${\frac{1second}{{10}^{-<!--\; -\; -->6}seconds}}^2$ inputs $n$, which I got from $\sqrt{n}microseconds=1second$ and solving for n.
My confusion comes from trying to make a proportion for the system like I would for a dimensional analysis problem. I have $\frac{\sqrt{n}microseconds}{ninputs}$ and the other fraction is $\frac{1second}{xinputs}$. Here when I solve for $x$ I get ${10}^6\ast <!--\; \ast \; -->\sqrt{n}$. I think my dimensions don't make sense somehow, but in my head, saying 'it takes square root of n microseconds for n inputs' comes out as $\frac{\sqrt{n}microseconds}{ninputs}$
Thank you!!

How does this separation of the variable from the fraction result in
$3\cdot <!--\; \cdot \; -->\frac{1}{x}$ & not $\frac{3}{x}$
I wish to write the following expression as the product of a whole number or a fraction and variable expression. The answer given in my textbook is as follows:
$\frac{3}{x}=\frac{3\cdot <!--\; \cdot \; -->1}{1\cdot <!--\; \cdot \; -->x}=\frac{3}{1}\cdot <!--\; \cdot \; -->\frac{1}{x}=3\cdot <!--\; \cdot \; -->\frac{1}{x}$
However, I am not sure how to get to $3\cdot <!--\; \cdot \; -->\frac{1}{x}$. When I complete the multiplication I get to $\frac{3}{1}\cdot <!--\; \cdot \; -->\frac{1}{x}=\frac{3}{x}$, which of course loops me back round!

Why **Excursion length** is less than **Denominator**?
Can anyone prove that the Excursion length of any fraction's decimal representation which is cyclic, is less than its Denominator?

Logic behind absolute value problem
I'm trying to nail down the logic behind absolute value inequalities. Given an inequality like
$|\frac{2}{x+4}|>2,$
the normal process is to use cases and the definition of absolute value to determine the solution set.
However, upon thinking about the process a little more I've confused myself. When faced with this problem the goal is to find the values of $x$ that make the relation true. But by using cases I am assuming that $\frac{2}{x+4}$ is nonnegative in one case and negative in another and then I proceed to find the solutions that satisfy each case. Is this logic circular since the sign of $\frac{2}{x+4}$ depends on $x$? It seems like I'm assuming something about $x$ before I find it. Can someone explain the logic here? Thanks for the help.

Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge <!--\; \ge \; -->3$
I have to prove that
$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge <!--\; \ge \; -->3$
is always true for real numbers $a,b,c>0$ with $abc=1$
Using the AM-GM inequality I got as far as
$\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ca}{1+c}\ge <!--\; \ge \; -->\frac{b}{\sqrt[a+1]{b}}+\frac{c}{\sqrt[b+1]{c}}+\frac{a}{\sqrt[c+1]{a}}$
but I do not yet know how to finish my proof from there (if this is helpful at all?!).