Given the inequality:  $\frac{1}{x} &lt; 5$  In order to find a solution, I would normally multiply both sides by $x$:  $1 &lt; 5x$  Then I would divide by $5$  $\frac{1}{5} &lt; x$  To obtain the solution: $x &gt; \frac{1}{5}$.  Now, the thing is, the solutions are actually two: $x &gt; \frac{1}{5}$ and $x &lt; 0$  How am I supposed to reach this conclusion algebraically? It seems I'm not able to obtain the second solution ($x &lt; 0$).  Thanks!
121303
Solving $\frac{1}{x} &lt; 4$ gives me $x &gt; \frac{1}{4}$. The book however states the answer is: $x &lt; 0$ or $x &gt; \frac{1}{4}$.  My questions are:  Why does this inequality has two answers (preferably the intuition behind it)?  When using Wolfram Alpha it gives me two answers, but when using $1 &lt; 4x$ it only gives me one answer. Aren't the two forms equivalent?
15535
Problem: Show that if $x_n\to 5$ then $\dfrac{x_n +1}{\sqrt{x_n -1}} \to 3$.  So I know $|x-5|&lt;\epsilon$ and I need to show that $\left|\dfrac{x_n +1}{\sqrt{x_n -1}} -3\right| &lt; \epsilon$.  I tried applying the triangle inequality (and a bit of algebra) to get $$\left|\frac{x_n +1}{\sqrt{x_n -1}} -3\right| \leq |x_n -5| \left |\frac{1}{\sqrt{x_n -1}}\right| + \left|\frac{6}{\sqrt{x_n-1}} -3\right|.$$  From this I know that $|x_n -5| &lt;\epsilon$ and can show that $\left |\dfrac{1}{\sqrt{x_n -1}}\right|&lt;1$. So my only problem is the second part of the sum. I have no idea what to do with it to get everything to be less than $\epsilon$.
122977
I have to prove that the result of $x * y \in G$ so $\frac{x+y}{1+xy} \in (-1, 1)$. So $x &gt; -1$ and $y &gt; -1$ at the same time $x &lt; 1$ and $y &lt; 1$.  If I multiply the first 2 expressions I obtain $xy &lt; 1$ which is true, and if I sum up the last 2 I get $x + y &lt; 2$. At this point I am not sure if I am doing well. How should I continue this problem?
101605
I am trying to prove the above statement but I'm not sure if my proof is correct. My proof is as follows,  Given $u\cdot v$, we know by the C-E Inequality that $|u \cdot v| \leq ||u|| \ ||v||$  Consider $\frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2$  We can manipulate this as follows $$\frac{1}{4}(||u+v||^2 - ||u-v||^2) = \frac{1}{4}(||u+v||- ||u-v||)(||u+v||+||u-v||)$$ Using the triangle inequality the above can be written as follows, $$...\leq \frac{1}{4}(||u|| +||v|| -||u||+||v||) (||u|| +||v|| +||u||-||v||) = \frac{1}{4} (4||u||||v||) = ||u||||v||$$  As such both $u\cdot v$ and $u\cdot v = \frac{1}{4}||u+v||^2 - \frac{1}{4}||u-v||^2 \forall u,v \in \mathbb{R^n}$ are $\leq ||u|| ||v||$  I'm not sure whether I've proven this correctly because I ended up proving that both sides of the equation are less than or equal to something entirely different (the magnitude of v multiplied by the magnitude of u). Is this acceptable? I don't think it is, since if both can be less than or equal to the magnitudes multiplied then they do not necessarily have to be equal to each other. If this is the case, can someone give me a hint on how to move forward with this proof?
118257
What is the meaning of $x$ raised to any non-positive value? We know that $x^{-a} = \dfrac{1}{x^a}$ and $x^0 = 1$, but where does that come from? What is the proof? Why is this true? What about $x$ raised to a fraction, like say $\frac{1}{3}$? How do you multiply $x$ $\frac{1}{3}$ times?
121268
I want to know what will happen to the inequality in these cases  $a&gt;x&gt;b$ If $a$ and $b$ negative  When i square the both sides of it  Is it going to be  $b^2&gt;x^2&gt;a^2$  Ok but what would happen if  $a$ is a positive and $b$ is a negative number when i square ? EXAMPLE $2&gt;x&gt;-1$  another case  $2&gt;x&gt;-4$   i forgot  mentioning this case ?  If it is something like  $2&gt;x&gt;-2$
96105
I have the following question:          Find the value of:     $$ \frac {1} {\sqrt {4} + \sqrt {5}} +  \frac {1} {\sqrt {5} + \sqrt {6}} + ... +  \frac {1} {\sqrt {80} + \sqrt {81}}$$      The book already provide the answer for this question but it didn't include the ways. Hint or anything that will help is appreciated
117645
Prove that the inequality $$\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|&lt;\frac{1}{mn}$$ holds for positive integers $m, n$ if and only if $m$ and $n$ where $m &gt; n$ are two successive terms of the Fibonacci sequence.   I thought about using the explicit formula for the Fibonacci sequence, which is $$F_n = \dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right),$$ but this seems to get computational. Is there an easier way to think about this?
34511
I came up some definitions I have sort of difficulty to distinguish. In parentheses are my questions.   $\dfrac {x}{0}$ is Impossible (  If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both $1.$ and $2.$ are divided by zero, but only $2. $ has infinite solutions so as $1.$ has none solution, how and why ?) $\dfrac {0}{0}$ is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?) $\dfrac {0}{x}$ and $x \ne 0$, it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it.
36693
I came across this example of how you could end up with an extraneous solution but I was wondering how it arose. We have the equation: $$x^2+x+1=0 $$ Since x=0 does not satisfy the equation, you can divide by x on both sides which yields: $$x+1+\frac{1}{x}=0$$ which is equivalent to our first equation. From our first eqution we can conclude that: $$-x^2=x+1$$ We now substitute that into the second equation to get:$$x^2=\frac{1}{x}$$ which results in $$x^3=1$$ which is equivalent to our previous equation since x cannot be 0. However, one solution from our last equation is x=1, which is not a solution to our original equation. I have a vague idea that it may have to do with the fact that you get a cubic equation and you began with a quadratic, and that steps imply the following and not vice versa, but can you provide a very detailed answer as to why it arises? Can you please provide more examples?
72180
This is a classic problem, but I haven't seen it on the site before.     Suppose $x$ and $y$ are real numbers that both lie in the interval   $(0,1)$. Prove  $$x^y+y^x&gt;1.$$   The following spoiler box contains a hint that indicates the solution I know. Does anyone know a more natural solution?       Use Bernoulli's inequality to show $$x^y\ge \frac{x}{x+y}.$$   Edit: I recently learned the source is the 1996 French math olympiad.
30048
Show that   $$\displaystyle1 - \frac{x^2}2\leq\cos x\quad\forall x \in \mathbb{R}$$   Let $f(x) = \cos x - 1 + \frac{x^2}2$; then we need to show that $f(x) \geq 0\quad\forall x \in \mathbb{R}$.  Since $$\lim_{x \to \pm\infty} f(x) = +\infty$$ and $$\operatorname{im}f'' = [0, 2]$$ (as $f''(x) = 1 -\cos x$), it remains to show that $x = 0$ is a minimum for $f(x)$, but I cannot do that because I would need to solve $f'(x) = x -\sin x = 0$, which I cannot. I think it would suffice to show that plugging in $x = 0$ we have that $f'(x) = 0$, and combining it with the fact that the function $f$ is convex, we would be done.    Is it enough? What's a better proof?  EDIT: It just occurred to me that there is a way to solve $f'(x) = 0$. Just observe that $f'(x)$ is an odd function and that $f''(x) \geq 0$ implies that $f'(x)$ is monotonically increasing. Since an odd function, if defined at $x = 0$, must be $0$ there, $x = 0$ is a solution to the equation and it's the only one.
76504
I'm struggling with proving the following inequality: $$\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}$$  for all $t \geq 2$. I think it is monotonic non-increasing in $t$, which would suffice.  Thanks in advance!
24417
I have the equation $x^2=x$.  If I divide $x$ from both sides I get $x=1$.  Yet clearly $x$ can also equal $0$.  What step in this process is wrong? It seems to me that there's only one step. And isn't dividing the same thing from both sides a valid step?  I hope this isn't a stupid question because I feel dumb asking about something so basic.  EDIT: To clarify, what I'm looking for is not only an explanation for why my methodology is wrong, but also a better methodology that will keep me from missing possible solutions in the future.
78664
I'd like to prove that $$\frac{(a-b)^2}{8a}\leqslant\frac{a+b}{2} - \sqrt{ab}\leqslant\frac{(a-b)^2}{8b}$$ when $ 0&lt;b\leqslant a\,$. I have been brushing up on inequalities and I cannot figure this one out. I have searched this site, and found this  , but haven't seen how to apply it. I have tried to find a factor that I can multiply the entire inequality with that will allow me to simplify the middle term. So far no luck there. I have also tried assuming that $b &gt; a$ and looked for a contradiction to either of the terms individually. Does anyone have any hints that can get me somewhere?
113910
EDIT:  My question was poorly worded.   I wasn't asking about showing $\ln(1+x) &gt; \frac{2x}{2+x}$ for $x&gt;0$.   What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for small positive values of $x$. This is addressed in robjohn's answer.
11088
How do I prove the following inequality  $$\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$$  without the help of induction?  Thanks for any help!!