,
,
,
,
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In ![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
be an acute-angled triangle with 
.
be the circumcircle of 
be the midpoint of the arc 
of 
.
meets 
at 
and meets 
.
at 
.
by 
.
.
meets line 
.
and 
with respect to side 
with the circle of diameter 
and 
are the two intersections of the perpendicular from 
with respect to side 
with the circle of diamter 
and 
lies on 
satisfying![\[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]](http://latex.artofproblemsolving.com/8/0/f/80fee0440e9fc4031fa4dcff96b9605011ef0e68.png)
for all real numbers 
and 
.
has how many digits?
Y by Adventure10 and 1 other user
Let 
be real numbers. Prove that
![\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]](//latex.artofproblemsolving.com/0/9/f/09fee86fb6ae5286ff38443f1590d912c4831d80.png)
Greece
EDIT BY DARIJ:
Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
be real numbers. Prove that![\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]](http://latex.artofproblemsolving.com/0/9/f/09fee86fb6ae5286ff38443f1590d912c4831d80.png)
Greece
EDIT BY DARIJ:
Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
This post has been edited 1 time. Last edited by Valentin Vornicu, May 3, 2006, 3:18 PM
be real numbers. Prove that ![\[ \frac 1{a(1+b)} + \frac 1{b(1+c)} + \frac 1{c(1+a)} \geq \frac 3 { 1+abc}. \]](http://latex.artofproblemsolving.com/e/c/e/eceb19d255038b777dc8cda4b96154efee7c590b.png)
.
where x is the best real .
where y is the best contsnt .![$[\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}][a(b+1)+b(c+1)+c(a+1)]\geq xy\geq 9$](http://latex.artofproblemsolving.com/3/2/f/32f31749d01431fa00a53ce2ac07882eac017b35.png)
and so 
and so
,
now could you please see my solution and tellme if it is correct ? i have a kind of battle with someone and i have to know from people like you if my solution is correct 
and 
.
and 
does not imply that 
.
are the "best" constants, you have 
,
for a certain constant 
would be the best constant, right? Which is equivalent to showing that there is always an equality case on the AM-GM for any 
so it only works for 
.
be positive real numbers . Prove that 
