Given $s \in (0,1)$, prove $$x^sy^{1-s} \leq sx + (1-s)y$$ for $x,y &gt; 0$  Tried some algebraic manipulations but I'm guessing I need to use some trick.  Any suggestions, hints?
I have no idea how to do this. Any help would be appreciated. The chapter I'm on is about differentiation and the mean value theorem.  Prove $a^\alpha b^{1-\alpha} \le \alpha a + (1 - \alpha)b, \; a,b &gt; 0,\; 0 &lt; \alpha &lt; 1$
How to get a toddler to start chewing and eating solids? My son is 18 months old and still doesn't want to chew and eat solid foods. We have to blend soups and other food into purée for him to accept it. If there is even a slight solid, he will automatically push it out of his mouth with his tongue.  I have had a breakthrough with breakfast where I don't blend as much and leave it a little chunky. He ate it but still pushed chunks of soft banana that were bigger out.  Some of my friends have kids that are 4-5 months younger, who are eating chicken cutlets.  Any ideas or proven methods for getting a child to start eating solids? How to stop the find command after second match? How to make find command, that will show only first two matches and then end, something similiar to  just version for two records. What focal length lens do I need for photographing the moon? Define $\sigma(m) = \sum$ d : d|n.  Prove that $p^j$$q^i$ cannot be a perfect number for $p, q$ odd, distinct primes.  Attempt at Solution: I have shown that $p^k$ can never be a perfect number, and im trying to use the multiplicative property of $\sigma$ to generalize to $p^j$$q^i$ Let $m$ be Lebesgue measure and $a \in R$. Suppose that $f : R \to R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. This is a corollary to a proof in Bass, but I don't understand why it follows from the proof he gives. I follow everything up until the last statement. Why is it that proving that the integral is $0$ for all Borel measurable sets guarantees it is true for all Lebesgue measurable sets? Am I missing something vital about the relationship between Lebesgue measurable sets and Borel measurable sets? Statement: Let $m$ be Lebesgue measure and $a \in\mathbb R$. Suppose that $f : \mathbb R \to\mathbb R$ is integrable, and $\int_a^xf(y) \, dy = 0$ for all $x$. Then $f = 0$ a.e. Proof: We prove the conclusion first for intervals, then for unions of intervals, then for open sets, then for Borel measurable sets. For any interval $(c,d)$, we have $$\int_c^df = \int_a^d f - \int^c_a f = 0.$$ By linearity, if $G$ is the finite union of disjoint open intervals, then $\int_G f = 0$. Now, if $G$ is open, then $G$ is the countable union of disjoint, open intervals, $G = \bigcup^n_{i=1}I_i$. We have $$\int_G f = \int \lim_{n\to\infty} f \cdot \chi_{\bigcup^n_{i=1} I_i}$$ and since $f_n$ is integrable for all $n$, and $|f_n| \le f$ for all $n$, by the dominated convergence theorem, $$\int_G f = \lim_{n\to\infty} \int_{\bigcup^n_{i=1}I_i} f =0.$$ Now, if $G_n$ is a decreasing sequence of open sets converging to $H$, since $|f \cdot \chi_{G_n}| \le f \cdot \chi_{G_1}$ for all $n$, we again have by the DCT that $\int_H f = 0 = \lim_{n\to\infty} \int_{G_n}f = 0$. Finally, if $G$ is a Borel measurable set, then we know that there exists a sequence $G_n$ of decreasing, open sets, that converge to some set $H$, where $H \setminus G$ is a null set. Thus, $$\int_G f = \int_H f = 0.$$ We have found that for every Borel measurable set $G$, $\int_G f =0$. Since $f$ is real valued and measurable, $f = 0$ a.e. If verification is quality control, what is a validation? Length of generator output Python provides a nice method for getting length of an eager iterable, len(x) that is. But I couldn't find anything similar for lazy iterables represented by generator comprehensions and functions. Of course, it is not hard to write something like:  def iterlen(x):   n = 0   try:     while True:       next(x)       n += 1   except StopIteration: pass   return n   But I can't get rid of a feeling that I'm reimplementing a bicycle.  (While I was typing the function, a thought struck my mind: maybe there really is no such function, because it "destroys" its argument. Not an issue for my case, though).  P.S.: concerning the first answers - yes, something like len(list(x)) would work too, but that drastically increases the usage of memory.  P.P.S.: re-checked... Disregard the P.S., seems I made a mistake while trying that, it works fine. Sorry for the trouble. Custom Button on Lightning Experience I created a custom button Send Email in Lightning Experience. The problem is that button in visible in feed section layout, but I want to put it in the upper right corner of page layouts near Edit, Clone, Delete ecc....  How Can I do that? Thanks  I add an image that explain the situation.    EDIT: I would like to explain you better my problem.  I have created three custom quick action:   I have added them to the case page Layout as you can see    I would like to understand the reason why I don’t see these three custom action in the Page Layout in the section in the top right Using zypper, how can I remove all packages installed from a specific repository? What antivirus program can I use on Ubuntu? Are there any antivirus applications that work on Ubuntu 16.04? Thanks! Can not switch Node.js to a default version of my choice I am trying to execute a web service call to our backend system when the opportunity is closed/won. I am attempting to do a  POST to our endpoint. All the api call is really carrying the account name and a few structural detail back (All listed below). My problem is that when the trigger executes I get the following error in the debug logs "System.CalloutException: You have uncommitted work pending. Please commit or rollback before calling out".  The Process Builder Passing in the company name:   APEX Class  public class Play2 { @TestVisible static HttpResponse res;  @InvocableMethod public static void sendNotification(List&lt;String&gt; name) {        HttpRequest req = new HttpRequest();     Http http = new Http();     String payload = JSON.serialize(         new Map&lt;String, String&gt; {             'companyName' =&gt; name.get(0),             'salesforceAccountType' =&gt; 'Company',             'salesforceSynchStatus' =&gt; 'Enabled'         }     );      req.setEndpoint('https://TheEndPoint');     req.setMethod('POST');     req.setHeader('Authorization', 'The Mighty API KEY');     req.setHeader('Content-Type', 'application/json');      req.setBody( payload );     req.setCompressed(true);       try {         System.debug('Point 1');         res = http.send(req);         System.debug('Point 2');     } catch(System.CalloutException e) {         System.debug('Callout error: '+ e);         System.debug(res.toString());     } } } Inequality $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ How do real and effective user IDs work? When a normal user wants to make change to the passwd file, the user will by setuid be given the effective user access. User becomes root temporarily and can edit passwd.  However you can only edit your password right, and not everybody else? However your effective user access is root. So how come you're not allowed to change other passwords beside yours?  When you run a program with setuid, what does it mean actually when effective user is root, but real user id is still your name? Impossibility of constructing a continuum-size linearly independent set in $\Bbb R$ This is a response to the following exchange at    [Bill constructs a $\aleph_0$ linearly independent set in $\Bbb R$]  Can you extend the idea of this proof to get the right dimension? Currently, you only have a countably infinite independent set, but the dimension is size continuum. –   I believe the answer to this question is no, for the following reason. Suppose we have a spanning set for $\Bbb R$ over $\Bbb Q$, of cardinality $\frak n$. Then by definition we have the cardinality equation ${\frak c}\le^*[{\frak n}]^{&lt;\omega}\cdot \aleph_0$ (where $[{\frak n}]^{&lt;\omega}$ is the cardinality of finite subsets, and $\frak m\le^*n$ means there is a surjection from $\frak n$ to $\frak m$). If $\frak n$ is finite, then we get ${\frak c}\le^*\aleph_0$ which is false, so $\frak n$ is infinite (but not necessarily Dedekind-infinite). If we assume the axiom of choice, then the equation reduces to $\frak c\le n$, hence every Hamel basis is size continuum. But without AC I don't know any way to simplify the original equation to get anything stronger than &quot;$\frak n$ is infinite&quot;. This is at best heuristic evidence for the original question, since it is considering a different problem (spanning sets instead of linearly independent sets). How could one show that there are no ZF-constructible $\Bbb Q$-linearly independent subsets of $\Bbb R$ of cardinality continuum? Find the value of $g'(0)$. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function with $f(1) = 5$ and $f(3) = 11$. If $g(x) = \int_1^3 f(x+t)dt$ then find $g'(0).$   Please explain your answer. I would like to show the author names using \citet but I get just the reference number. Here is a minimal code:  \documentclass{report} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{natbib}  \begin{document}   The author \citet{toschi09} has shown that ..  \bibliographystyle{plain} \bibliography{foo}  \end{document}   And here is foo.bib  @article{toschi09, author = {Toschi, F. and Bodenschatz, E.}, title = {Lagrangian Properties of Particles in Turbulence}, journal = {Annual Review of Fluid Mechanics}, volume = {41}, number = {1}, pages = {375-404}, year = {2009},  }   What I get is:     The author (author?) [1] has shown that ..   Any help will be more than welcome, thank you! Recovering data from a faulty disk When I plug in a removable 1TB USB disk in my laptop, the operations led keeps blinking for a while and then you can hear the servo mechanism resetting with a short "tack".  I'd like to recover the data.    I believe there might be some way to get the data through low level IOs.    What I believe is that the mounting process assumes the disk is OK and I should prevent this before using ddrescue or something similar.  Does anybody have a solution to this problem? Is there a canonical explanation for how Ant-Man's weight works when he shrinks down that's consistent with the various events depicted in the film?    I think the explanation given in the film is roughly "the suit compresses the space in-between atoms, so your size decreases, your density increases, and your total mass stays the same", however it seems to be portrayed in contradictory ways.  For instance:   He can float in water, and even be sent shooting through the air by a relatively gentle flow of water filling a bathtub.  This wouldn't happen if he retained his original mass, as he'd essentially be the densest thing on the planet and should sink like a stone even in a vat of molten lead. Upon landing after being ejected from the bathtub, the ceramic tile underneath him shatters.  That takes a significant amount of force, and similar effects are used at other points in the movie where he falls onto (or straight through) various things. He's able to get sucked into a vacuum-cleaner.  One that didn't even appear to be a high-end, extra powerful vacuum-cleaner, let alone one capable of sucking up a 200 pound piece of dust.   He can run along the barrel of a gun as someone shoots it, apparently without causing them to drop the weapon or even have trouble continuing to aim it.     Does a more consistent explanation exist in other media, that would fit with the above examples?  Or is this simply a case of the power being made to do whatever happened to be most convenient at the time? Why does the dock appear on the left when my screen is locked? The dock doesn't work correctly after I enter my password, either.