Explore Fraction Concepts with Examples and Questions

Inequality $\frac{3}{\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}}\ge <!--\; \ge \; -->1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ for positive $a$
MyTry::We can write $A$ and $B$ as
$=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{3}{A}$
and
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{B}$
Using cauchyschwarz Inequality
$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge <!--\; \ge \; -->\frac{3^2}{1+a+1+b+1+c}=\frac{9}{3+a+b+c}$
Now How can i solve after that , Help Required, Thanks

Solving for x [not homework]
How do I bring the remaining x to the LHS?
$\pm <!--\; \pm \; -->x=\frac{(2-<!--\; -\; -->x)\sqrt{|q_2|}}{\sqrt{|q_1|}}$
to get
$x=\frac{2\sqrt{|q_2|}}{\sqrt{|q_2|}\pm <!--\; \pm \; -->\sqrt{|q_1|}}$
I'm just not sure about the background operations to isolate the x.

Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$is not in the lowest term?
I found $n=2$ be a solution. Is it correct?

Derivative of $f(x)=\frac{\sqrt{x^2-<!--\; -\; -->1}}{x}$
I have the function following:
$f(x)=\frac{\sqrt{x^2-<!--\; -\; -->1}}{x}$
And here is what I did:
$f(x)=\frac{\sqrt{x^2-<!--\; -\; -->1}}{x}$
$=\frac{(x^2-<!--\; -\; -->1{)}^{\frac{1}{2}}}{x}$
$f^{\prime }(x)=\frac{x\cdot <!--\; \cdot \; -->\frac{1}{2}(x^2-<!--\; -\; -->1{)}^{-<!--\; -\; -->\frac{1}{2}}(2x)-<!--\; -\; -->(x^2-<!--\; -\; -->1{)}^{\frac{1}{2}}}{x^2}$
$=\frac{x^2(x^2-<!--\; -\; -->1{)}^{-<!--\; -\; -->\frac{1}{2}}-<!--\; -\; -->(x^2-<!--\; -\; -->1{)}^{\frac{1}{2}}}{x^2}$
And I'm pretty sure that this is wrong, and the answer book says it isn't either.
I think I messed up somewhere, or didn't do it properly.
I tried using the quotient rule. Do I need to make it into an exponent, and solve it as a chainrule?
Please help me find the steps and answer to this question. Thank you.
Or is there any other steps so it can match this?:
$\frac{1}{x^2\sqrt{x^2-<!--\; -\; -->1}}$

inequality with three variables and condition
If $a$, $b$ and $c$ positive real numbers such that $a+b+c=1$, prove
${\displaystyle \frac{b^2}{a+b^2}}+{\displaystyle \frac{c^2}{b+c^2}}+{\displaystyle \frac{a^2}{c+a^2}}⩾<!--\; ⩾\; -->{\displaystyle \frac{3}{4}}$. I have tried several methods to solve this,but can't get any result. Any idea?

Confusing sum of fractions
Question is to find the sum of:
$(\frac{1}{2^2-<!--\; -\; -->1})+(\frac{1}{4^2-<!--\; -\; -->1})+(\frac{1}{6^2-<!--\; -\; -->1})+(\frac{1}{{20}^2-<!--\; -\; -->1})$
I know that $a^2-<!--\; -\; -->b^2=(a+b)(a-<!--\; -\; -->b)$ , and that with this I can find the LCM to be 1995, but this isn't helping me find the answer. I am looking for some kind of formula, because this question can come in an exam (having 60 qs for 60 mins), so they must not be giving a brute-force approach question.
Please help me in finding a fast approach.
Thanks

Simplifying Fractions involving negative numbers
I want to simplify
$\frac{\frac{7}{-<!--\; -\; -->10}\times <!--\; \times \; -->\frac{-<!--\; -\; -->15}{6}}{\frac{7}{-<!--\; -\; -->19}+\frac{-<!--\; -\; -->17}{-<!--\; -\; -->8}}$
I really don't understand how to do this, or even how to start? Negative numbers make it even harder for me to try and simplify it.

Find equation for mass in gravity
A satellite is moving in circular motion round a planet.
From the physics we know that
$\mathrm{\Sigma <!--\; \Sigma \; -->}{F}_r=m{a}_r=\frac{GMm}{r^2}$
So I wanted to find the equation for $M$ knowing also that
$v=\mathrm{\omega <!--\; \omega \; -->}r=\frac{2\mathrm{\pi <!--\; \pi \; -->}r}{T}$
and
$a_r=\frac{v^2}{r}$
Thus,
$m{a}_r=\frac{GMm}{r^2}$
$a_r=\frac{GM}{r^2}$
$\frac{v^2}{r}=\frac{GM}{r^2}$
$\frac{{\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}r}{T}\right)}^2}{r}=\frac{GM}{r^2}$
$\frac{\frac{4{\mathrm{\pi <!--\; \pi \; -->}}^2{r}^2}{T^2}}{r}=\frac{GM}{r^2}$
$\frac{4{\mathrm{\pi <!--\; \pi \; -->}}^2{r}^3}{T^2}=\frac{GM}{r^2}$
$\frac{4{\mathrm{\pi <!--\; \pi \; -->}}^2{r}^5}{T^2}=GM$
$\frac{4{\mathrm{\pi <!--\; \pi \; -->}}^2{r}^{5}G}{T^2}=M$
However, this is wrong! It should be:
$M=\frac{4{\mathrm{\pi <!--\; \pi \; -->}}^2{r}^3}{G{T}^2}$
What was my mistake in Mathematics? Please don't migrate it to physics because my misunderstanding is on math.
Note: I would be very happy if you show my mistake, instead of showing me another way to get to the equation.

decimal to fractions
When being asked how to solve the Arithmetic Means of 8,7,7,5,3,2, and 2, I understand that adding these numbers then dividing by 7 (the amount of numbers) gives me the decimal $\mathrm{4.85714...}$
But when being asked to change this number into a fraction, I do not understand where the $\frac{6}{7}$ comes from in the answer of $4\frac{6}{7}$. Can someone please help me understand?

Find $\frac{y}{x}$ from $3x+3y=yt=xt+2.5x$ I need to find the ratio of
$\frac{y}{x}$
If given that
$3x+3y=yt=xt+2.5x$
So what I tried is:
$t=\frac{3x+3y}{y}$
And then put it in the equation
$\frac{x(3x+3y)}{y}+2.5x=\frac{(3x+3y)}{y}y$
$\frac{x(3x+3y)}{y}+2.5x=3x+3y$
$\frac{3x^2}{y}+\frac{3yx}{y}+2.5x=3x+3y$
$\frac{3x^2}{y}+3x+2.5x=3x+3y$
$\frac{3x^2}{y}+2.5x=3y$
Here I got stuck. I didn't know how to find the ratio. Can someone help me?

What happens when you add $x$ to $\frac{1}{3}x$?
I am dealing with an equation that requires me to add $x$ to $\frac{1}{3}x$:
$x+\frac{1}{3}x$ = ??
I know this might be simple to any of you on this site, because you are all asking questions with symbols I have never seen, but this is confusing to me.
I guess one way of thinking about this is - You are adding $x$ to $yx$, right? Or just adding another $\frac{1}{3}$?
The complete equation that I am working on is [- don't laugh at its simplicity ;)]:
$\frac{2}{3}b+5=20-<!--\; -\; -->b$
So, when worked out... I got:
$\frac{2}{3}b+b=15$
And this is where I get stuck.

Hard problem with fractions
I can't solve the following problem.
A person is x years old. Find his age if the following is true. In a group of x people each one started taking pictures of each of the others. At some point we know that more than $\frac{1}{2}$ of the people has taken exactly $\frac{1}{2}$ of all of their pictures (which are $x-<!--\; -\; -->1$ for every one of them), at the same time more than $\frac{1}{3}$ of them has taken exactly $\frac{1}{3}$ of all of their pictures, and at the same time more than $\frac{1}{7}$ of them has taken exactly $\frac{1}{7}$ of their pictures.

Euler's Continued Fraction Theorem for fractions
How can I use Euler's Continued Fraction Theorem to find the continued fraction expansion for a (ordinary, finite) fraction via its (terminating or recurring) decimal expansion, rather than via the more obvious Euclid's Algorithm?
The examples I'm thinking of are $5/7$ and $3/8$ each of which can be thought of as a series in a power of $10$
$5/7=0.\stackrel{¯<!--\; ¯\; -->}{571428}=\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}571428({10}^{-<!--\; -\; -->6}{)}^k$
and
$3/8=0.375=3\ast <!--\; \ast \; -->{10}^{-<!--\; -\; -->1}+7\ast <!--\; \ast \; -->{10}^{-<!--\; -\; -->2}+5\ast <!--\; \ast \; -->{10}^{-<!--\; -\; -->3}.$
I'd like to use these coefficients and expansion points together with Euler's Theorem to show the continued fraction expansions
$5/7=[0;1,2,2]$
and
$3/8=[0;2,1,2].$

How does this numerators and denominators relate with the fraction?
Suppose if we have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ then how are their values related with the fraction $\frac{a+c}{b+d}$?
I have observed this inequality: $\frac{a}{b}\le <!--\; \le \; -->\frac{a+c}{b+d}\le <!--\; \le \; -->\frac{c}{d}$
Does this hold true $\mathrm{\forall <!--\; \forall \; -->}a,b,c,d\in <!--\; \in \; -->N$ ? Is there any proof?
I am sorry if this question is too basic, but I want help regarding this.

variables to the power of a fraction
I have this question for advanced math, I can't seem to get my head around.
$\frac{x^{5/2}}{(x^{1/3}{)}^4}$

Fraction with negative exponent fraction.
Q:
${\left(\frac{27a^6{b}^{-<!--\; -\; -->3}}{c^{-<!--\; -\; -->2}}\right)}^{-<!--\; -\; -->2/3}$
A:
$\frac{b^2}{9a^4{c}^{4/3}}$
How in the world are they getting that?

How can I shorten this expression? $\frac{-<!--\; -\; -->x\cdot <!--\; \cdot \; -->a_1-<!--\; -\; -->x\cdot <!--\; \cdot \; -->a_2-<!--\; -\; -->x\cdot <!--\; \cdot \; -->a_3-<!--\; -\; -->u}{a_4}\cdot <!--\; \cdot \; -->l$
You see, there are many $-<!--\; -\; -->x$ terms. How can I make this shorter? Can I write:
$x\frac{-<!--\; -\; -->a_1-<!--\; -\; -->a_2-<!--\; -\; -->a_3}{a_4}\cdot <!--\; \cdot \; -->l-<!--\; -\; -->u\frac{l}{a_4}$
Is this right?
Regards.

Certain Fraction between Fractions
Is there always a fraction $\frac{r}{s}$ with $\frac{p}{q}<\frac{r}{s}<\frac{p+1}{q}$ and $s<q$ for $0<p<q-<!--\; -\; -->1\in <!--\; \in \; -->\mathbb{Z}$ and $r,s\in <!--\; \in \; -->\mathbb{Z}$?

Algebra Fraction Problem - Variable
I honestly can't figure out how to get this answer, I feel like a complete idiot. I've taken up to Calc 2, so I'm not an idiot just been a while.
I have the problem..
$(n-<!--\; -\; -->1{)}^2-<!--\; -\; -->\frac{(n-<!--\; -\; -->2)(n-<!--\; -\; -->1)}{2}$
The answer is..
$\frac{(n-<!--\; -\; -->1)n}{2}$
How do I get there??
$2\frac{(n-<!--\; -\; -->1)(n-<!--\; -\; -->1)}{2}-<!--\; -\; -->\frac{(n-<!--\; -\; -->2)(n-<!--\; -\; -->1)}{2}\dots <!--\; \dots \; -->$
I'm so lost..

"Canceling out" in division doesn't always work the same way does it?
I've been working on Nested Fractions at the Khan Academy. Recently I was doing a routine problem and came to the correct conclusion but I realized I didn't understand why I wouldn't keep dividing. Here's what I begin with. I'm supposed to reduce it to an equivalent simplified expression. I wish I could link to the problem but I had moved on after posting here. Here's the start:
${\displaystyle \frac{(1+\frac{x}{y})}{\frac{x}{y}}}$
Part way through my problem now looks like:
${\displaystyle \frac{y\cdot <!--\; \cdot \; -->(y+x)}{y(x)}}$
I decide to "cancel out" the y's out side the parentheses. I end up with...
${\displaystyle \frac{y+x}{x}}$
I understood this the be the right answer. But for a moment I was very tempted to cancel the remaining x's and be left with a y. Even though I do this canceling operation all the time I realized then that there is something about it I don't understand. So here I am. Can you explain why it is that I was able to cancel the y earlier but not the x near the end?
Much appreciated.