Take $x,y \in (0,\infty), a \in (1, \infty)$. For $a \in \mathbb N $ by binomial theorem it holds $$(x+y)^a \geq x^a + y^a.$$ Trying with some numbers it seems true also for $a \in (1, \infty)$. How can I prove it?
122653
Let $x, \ y, \ p$ be any real numbers with $x&gt;0$, $y&gt;0$, and $p&gt;1$.   The question is about (most probably) an elementary inequality:  Is it always true that  $x^p+y^p\leq (x+y)^p$ ?  Note that if $p$ is any positive integer, then the above inequality is obviously correct. What about if the number $p \ (\text{with} \ p&gt;1)$ is any non-integer real number?  I guess that (by my intuition) the answer should be positive. But how can we proceed to prove this inequality?
47364
Note: I don't want to get the full solution, but only a hint.  I have to show that for $x \in [0, \infty)$, the sequence $\left(1+\frac xk\right)^k$ is monotonically increasing. We were already given the hint that we could use the inequality of arithmetic and geometric means, but I don't see how to apply it yet.   I tried to do it by induction. While the case $k = 1$ works easily, I don't know how to start further from here. I already tried to use the definition of the binomial theorem, but it didn't lead me anywhere.
44043
We let a>0 and b>0 and  we have $\sum_{n=0}^{\infty}b^n-a^n$. We have to find all values a,b>0 where $\sum_{n=0}^{\infty}b^n-a^n$ is convergent. I think I can use that $$\sum_{n=0}^{\infty} b_n &lt; \infty , \sum_{n=0}^{\infty} a_n &lt; \infty \implies \sum_{n=0}^{\infty} (b_n-a_n) &lt; \infty. $$. And I think $\sum_{n=0}^{\infty}b^n$ and $\sum_{n=0}^{\infty}a^n$ is convergent if a&lt;1 and b&lt;1? It will make sense. But how can I show it?
116157
If $n\geq m$ show that: $(x^m+y^m)^{1/m} \ge (x^n+y^n)^{1/n}$, all numbers being positive real.   Obviously since $n\geq m$, for every $a\in \Bbb Z^+$: $a^{1/m}\geq a^{1/n}$, but does it help?
72318
I have not been able to solve this problem. Any insights would be appreciated!   Let $x, n > 1$ be integers. Suppose that for each $k > 1$ there exists an integer $a_{k}$ such that $x − a_k^n$ is divisible by $k$. Prove that $x = A^{n}$ for some integer $A$.
120377
Let $p$ be a prime. How do you show that the binomial coefficients $\binom{p}{i}$ are divisible by $p$ for $1\leq i\leq p-1$? And how does this result in the congruency $(x+y)^p\equiv x^p+y^p\pmod p$ for $x,y\in\mathbb{Z}$?
65397
Let $x&gt;0$. Find $$\lim_{n\to\infty}\frac{n^{x}}{(1 + x)^{n}}.$$   What I was thinking involved an inequality, by reason of considering the fact that  $$(1 + x)^{n} &gt; C(n, k) x^{k}$$ for some $k$ such that $n&gt;k$. Expanding the binomial coefficient and taking the $n$ common from each bracket, we have something like $$(1 + x)^{n} &gt; C(n, k) x^{n} &gt; \frac {x^{k}n^{k}}{k! 2^{k}}.$$ But now, I am totally clueless. By intuition, I know that the limit has to be zero, so I just want to prove using the concept of inequalities that what I thought was the right approach (being the sandwich theorem). Any help would be appreciated!
111542
$x,y,z \in \mathbb R$ My attempt: Notice that if $x^2 + y^2 + z^2 \geq xyz$ and $x+y+z \geq xyz$. So : $$x^2 + y^2 + z^2 \geq x+y+z $$ But this is actually not always true, take the case when $ 0&lt;x,y,z &lt;1$ And you will see that this is not working. Maybe I’m wrong.
115817
I am curious about whether $|x|^p$ with $0&lt;p&lt;1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$.   So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?
16105
How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$? I tried to identify the sum as a binomial series, but the $4$ and the $-1/2$ puzzle me. (This series arises in studying the first passage time of a simple random walk.)
6481
I would appreciate if somebody could help me with the following problem  Let $|x|\leq1, |y|\leq 1$. Show that  $$ 0  \leq x^{2}+y^{2} -2x^{2} y^{2}+2xy  \sqrt{1-x^{2}}  \sqrt{1-y^{2}} \leq 1 $$
123884
Given $a_n, b_n$ such that $\exists N   \forall n&gt;N  a_n, b_n \geq 0$ I want to show that if $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge then $\sum_{n=0}^{\infty}a_n b_n$ also converges.  I tried to show it in the following way, if $\sum_{n=0}^{\infty}b_n$ converges, then $\lim_{n\rightarrow\infty}b_n=0$, so there exists $n_0$ such that $b_{n_0}\geq b_n$ for all $n\gt n_0$, therefore $a_n b_{n_0}\geq a_n b_n$, so that $\sum_{n=0}^{\infty}a_n b_n\leq \sum_{n=0}^{n_0}a_n b_n+b_{n_0}\sum_{n=n_0+1}^\infty a_n$, and as my series is bounded by sum of finitely many terms and tail of convergent series multiplied by a constant it's also convergent. Is my proof correct, or are there any flaws in my reasoning?
50868
I am trying to prove $\exp(x+y) = \exp(x) \exp(y)$.  I may use that $$\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$$ I further know how to multiply two power series in one point, i.e. if $f(x) = \sum_{n=0}^\infty c_n(x-a)^n$ and $g(x) = \sum_{k=0}^\infty d_n(x-a)^n$ then $$ f(x)g(x) = \sum_{n=0}^\infty e_n(x-a)^n $$ with $$ e_n = \sum_{m=0}^n c_md_{n-m} $$
90172
if $A_n \longrightarrow \infty $ and $B_n \longrightarrow \infty $  then $(A_n+B_n) \longrightarrow  \infty$.  How do you prove it?
80322
Prove: $$ (a+b)^\frac{1}{n} \le a^\frac{1}{n} + b^\frac{1}{n}, \qquad \forall n \in \mathbb{N} $$ I have have tried using the triangle inequlity $ |a + b| \le |a| + |b| $,  without any success.
111590
I am trying to prove that $\binom{2n}{n} &lt; \frac{4^n}{\sqrt{2n}}$. I tried induction, but with no effect (all I can get to is $(2n+1)(2n+2) &lt; 4\sqrt{n(n+1)}$ which is false)
73031
Let $(a_n)_{n \in \mathbb{N}}$ a sequence of real numbers with $0 \leq a_n \leq 1$ for all $n \in \mathbb{N}$.  I want to prove the following inequality using mathematical induction:  $\prod \limits_{k=0}^{n}(1-a_k) \geq  1- \sum\limits_{k=0}^{n}a_k$.  It's obviously true for $n=0$.  $n \rightarrow n+1$  Assume that it's true for n. Then one has:  $\prod \limits_{k=0}^{n+1}(1-a_k) = (1- a_{n+1}) \prod \limits_{k=0}^{n}(1-a_k) \geq (1- a_{n+1}) (1- \sum\limits_{k=0}^{n}a_k) = 1- \sum\limits_{k=0}^{n}a_k - a_{n+1} + a_{n+1} \sum\limits_{k=0}^{n}a_k = 1- \sum\limits_{k=0}^{n+1}a_k + a_{n+1} \sum\limits_{k=0}^{n}a_k=...$  How to go on?