Let $x,y$ be positive real numbers with $1&lt;x&lt;y$ and $x^y = y^x$. Prove ${\bf carefully}$ that $x&lt;e&lt;y$.  My attempt to the solution. Notice that if $x^y = y^x$ we can write it as $\dfrac{\ln x }{x} = \dfrac{ \ln y }{y} $ because $x,y$ are both more than 1.We know that the function $f(t) = \dfrac{ \ln t }{t}$ is maximal at $t = e$. Thus, $\dfrac{ \ln y}{y} \leq \dfrac{ \ln e }{e} = \dfrac{1}{e} $ and so $$y \geq e \ln y $$ Now, if $e \geq y$, then $\ln y \leq 1$. I can't seem to see a contradiction from here. Any hint?? Next, Notice that $\dfrac{\ln x }{x} &lt; \dfrac{1}{e}$ then clearly $\boxed{x&lt;e}$. But, I am still stuck on showing $y&gt;e$.
18865
We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
99756
If $a$ and $b$ are positive real numbers such that $a+b=1$, prove that    $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\frac{1}{b}\bigg)^2\ge \dfrac{25}{2}.$$     My work: $$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge \dfrac{25}{2}\implies a^2+\dfrac{1}{a^2}+b^2+\dfrac{1}{b^2}+4\ge \dfrac{25}{2}$$ Now, we have $a^2+\dfrac{1}{a^2}\ge 2$ and $b^2+\dfrac{1}{b^2}\ge 2$. Here, I am stuck, I cannot use the information provided, $a+b=1$ to any use. Please help!
73806
I'm reading a book about proofs and I'm currently stuck in this problem.  Prove that for all real numbers $x$ and $y$ we have that: $$\lfloor x\rfloor \lfloor y\rfloor \leq \lfloor xy\rfloor \leq \lfloor x\rfloor \lfloor y \rfloor + \lfloor x \rfloor + \lfloor y \rfloor$$  I though I could do it by cases, considering when the number is a pure integer and when is an integer plus some real number. But by doing this I end up with having a lot of cases to show. Is there any better, simpler and clever approach? Thank you!  PS: I end up with a lot of cases because there is a point where I will have to consider "subcases" of cases specifically when the integer part is multiplying with the positive "rest" less than one
69535
Let $f(x) = |x|$ for $x \in [-1,1]$ and extend $f$ to all of $\mathbb R$ by requiring $f(x+2)=f(x)$. Prove that $|f(x)-f(y)|\le|x-y|$ for all $x$ and $y$.   This is easy to observe on a graph but my proof is long and a bit tedious. Just curious if there is a quick/elegant way?
16963
How to prove that the following definitions of $e^x$ are equivalent, with as simple tools as possible and without any knowledge of $e$ or logarithms ?  $$\sum_{n=0}^{\infty} \frac{x^n}{n!}=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$  Also preferably, prove that this is $a^x$ for some real number $a&gt;0$.
102433
This is a pretty straightforward question. I want to show $(x+y)^a &gt; x^a + y^a$ for $x,y&gt;0$ and $a&gt;1$.  One way would be this. WLOG, suppose $x \leq y$. Then:  $(1+\frac{x}{y})^a &gt;1+\frac{x}{y} \geq 1+(\frac{x}{y})^a$.  Noting that $\frac{1}{y^a}t_1&gt;\frac{1}{y^a}t_2 \implies t_1 &gt; t_2$, it follows that $(x+y)^a &gt; x^a + y^a$.  But what other ways are there to show this inequality?
19432
Let $x,y \in L_p$   such that  $\|x\|_p=\|y\|_p=1$ , $1&lt; p&lt;\infty$ and $x\neq y.$ Why is $\|x+y\|_p&lt;2$  ?  I'm not sure how to start the proof.. I don't know how to handle integral of $(x+y)^p$ and it seems that using the binomial theorem won't be a great success.
18326
Let $a, b$ be positive real numbers. Let $0 &lt; x &lt; 1$. Show that  $(a+b)^x &lt; a^x + b^x$.  The function $f(x) = (a+b)^x -a^x - b^x$ has $f(0) = -1$, $f(1) = 0$ and it seems to be an increasing function. But I am unable to show that it is increasing function.
77014
Why must the base of a logarithm be a positive real number not equal to 1? and why must $x$ be positive?  Thanks.
43654
Does anyone provide proof  of $1+x\leq e^x$ for all $x$?  What is the minimum $a(&gt;0)$  such that $1+x\leq a^x$ for all x?
71323
Prove: $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}$$ for all x, y positive   $$\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0$$  $$\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  $$\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0$$  All the elements are positive and if $\sqrt{x}=\sqrt{y}$ we get $0$  Is the proof valid?
117914
I was looking for a proof of $e^{x+y}=e^xe^y$ using the fact that $$e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n.$$  So I have that $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{(x+y)^k}{n^k}=\sum_{k=0}^n\frac{1}{n^k}\sum_{i=0}^k\binom{k}{i}x^iy^{k-i}=\sum_{k=0}^n\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\left(\frac{x}{n}\right)^i\left(\frac{y}{n}\right)^{k-i}$$  But I can't get $$\left(1+\frac{x+y}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\left(\frac{x}{n}\right)^k\sum_{i=0}^n\binom{n}{i}\left(\frac{y}{n}\right)^i.$$  Any idea ?
20631
Prove $(|x| + |y|)^p \le |x|^p + |y|^p$ for $x,y \in \mathbb R$ and $p \in (0,1]$.  A hint is given that for $0 \le \alpha \le \beta$ there exist a number $\xi \in (\beta, \alpha + \beta)$ s.t. $$(\alpha + \beta)^p - \beta^p = p \xi^{p-1}\alpha \le \alpha^{p}$$.  I see that by proving the above hint the result holds for positive numbers.  In order to prove the above hint I've considered the Intermediate value theorem, however I've not yet succeeded.
13721
$$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$  Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?
123333
These two are supposed to be equivalent because of the properties of logarithms, but the domains of $\ln(x^2)$ and $2\ln x$ seem different to me. For example, if I substitute $x=-1$ into the first, I get 0. But in the second, I get a non-real answer.   Why is this? The domains of these functions, when graphed, seem different as well. But everything I've been taught so far, and most of the things I can find on the web, do not explain this inconsistency. Typically, I consider the property $\ln(a^b) = b \ln a$ to be true...  I am only a senior in high school, currently in Calculus I, but hopefully the explanation won't be too outside of the realm of my understanding. But even if it is, I would still like to know the answer.   Thanks in advance
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?
67814
Prove $$\left(x^2 - y^2\right)\left(\frac1y - \frac1x\right) \ge 0$$ where $x$ and $y$ are positive real numbers.   Can I simplify $\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$ and then $(x^2 - y^2)(x-y) \ge 0$ cancelling out the $xy$?  Is this valid because then  \begin{align} (x+y)(x-y)(x-y)&amp;\ge  0\cdot(x-y)^2\ge 0 \end{align}  which is always true and then prove backwards from here?