Prove that if $x_n$ goes to x, then $(x_1x_2\ldots x_n)^{\frac{1}{n}}$ goes to $x$.
109401
This is not a duplicate of this .  The linked question says that it suffices to show that if $(x_n)\to x$ then $(\frac{x_1+\cdots+x_n}{n})\to x$ to prove my question, but how so? I tried using the same strategy as how one proves that if $(x_n)\to x$ then $(\frac{x_1+\cdots+x_n}{n})\to x$, by "splitting" the product in th $N$th term:  $$\sqrt[n]{x_1x_2\cdots x_n}=\sqrt[n]{x_1x_2\cdots x_Nx_{N+1}\cdots x_n}=\sqrt[n]{x_1x_2\cdots x_N} \sqrt[n]{x_{N+1}\cdots x_n}$$  but it seems I can't use for now the definition of convergence of $(x_n)$ because of the $n$th root. I also tried to use a result: if $(x_n)\to x$ then $(\frac{x_n}{n})\to 1$ but don't know if this is true. Sadly I've had no real progress. Any help will be greatly appreciated, thanks in advance!
109975
Let $a&gt;0$,$b&gt;0$ and $c&gt;0$. Prove that:    $$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$     $$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$  $\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$  $\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$  $\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$  I can't continue. Help me
89936
A polynomial $P$ with integral coefficients satisfies $P(n)&gt;n$ for all positive integers $n$. Every positive integer $m$ is a factor of some number of the form $P(1),\, P(P(1)),\, P(P(P(1))),\dots $. Prove that $P(x)=x+1$.
5503
I'm currently trying to prove the following:    And this is what I've come up with so far:    And now I'm stuck. Perhaps the next step might be to eliminate for all x?  I'm sorry for using links, but it seems like I'm not allowed to do it otherwise due to needed reputation point.  Help would be appreciated.
117774
$x\in \mathbb R\;\;$ such that $\;\;x+\frac{1}{x} \in \mathbb Z.$   Prove by induction: $$x+\frac{1}{x} \in \mathbb Z\implies x^n+\frac{1}{x^n} \in \mathbb Z, \;\;\;\forall n\in \mathbb N$$  I used the $complete/strong$ induction.  The base $n=1$ is already in the statement.  Assumption: Let $P(n)\ldots,\;x^n+\frac{1}{x^n} \in \mathbb Z$ be true, as well as $$P(1), P(2),...P(n-2),P(n-1).$$ Step:$\;P(n+1)$ is true.  $x+\frac{1}{x} \in \mathbb Z\;\wedge x^n+\frac{1}{x^n} \in \mathbb Z\implies \Big(x+\frac{1}{x}\Big)\Big(x^n+\frac{1}{x^n}\Big)\in \mathbb Z$   $x^{n+1}+\frac{1}{x^{n-1}}+x^{n-1}+\frac{1}{x^{n+1}}=\underbrace{\Big(x^{n-1}+\frac{1}{x^{n-1}}\Big)}_{\in\mathbb Z}+\Big(x^{n+1}+\frac{1}{x^{n+1}}\Big)$  $x^{n+1}+\frac{1}{x^{n+1}}=\Big(x+\frac{1}{x}\Big)\Big(x^n+\frac{1}{x^n}\Big)-{\Big(x^{n-1}+\frac{1}{x^{n-1}}\Big)}\in\mathbb Z$
9609
Show that if $n \geq 0$ and $x&gt;0$, then   $$ e^x &gt; 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}.$$  Not sure where to get started with this induction proof.
111764
Let $x_1= a(&gt;0) , x_{n+1} = x_n+\frac{1}{x_n}$ for $n \in \mathbb{N}$. Determine whether $(x_n)$ converges or diverges.   We have $x_1= a &gt; 0, x_2= a+\dfrac{1}{a},  (\dfrac{1}{a} &gt;0)$  which implies that $x_2 &gt; x_1$  Then assume for induction: $x_k &lt; x_{k+1} &lt; a+1 $ (since $\dfrac{1}{x_k} &lt; 1$)  $$x_k + \frac{1}{x_k} &lt; x_{k+1}+ \frac{1}{x_{k+1}} &lt; a + 1$$  $$\implies x_{k+1} &lt; x_{k+2} &lt; a + 1$$  This shows that that $x_k &lt; x_{k+1} \Rightarrow  x_{k+1} &lt; x_{k+2}$. Therefore $x_n &lt; x_{n+1}$ for all $n \in  \mathbb N$.  It follows by the monotone convergence theorem that $x_n$ is increasing and bounded.
115924
Prove that $$\frac{d}{dx}(\tan^{-1}(x))=\frac{1}{1+x^2}$$
3091
This is a follow-on from my  about samples from a distribution.  Suppose $X_1 \ldots X_{n-1}, X_n$ are random variables all following some fixed distribution $D$. How do I prove that $P(X_n &gt; X_1 \wedge \ldots \wedge X_n &gt; X_{n-1}) = \frac{1}{n}$ (this is true intuitively and experimentally).
25938
$\dfrac{n}{2} \le \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n - 1} \le n $  I've Tried for hours but didn't got any striking idea. I don't have any efforts to show rather than induction. Please Try If you 've non-inductive proof.
3540
I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$
75449
Just would like to see if I'm right.  Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction.   Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$   Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.  Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.  I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
106369
Prove that for $x&gt;0$  and $0&lt;p&lt;q$:  $$\left( 1+ \frac{x}{p} \right) ^{p}  &lt; \left( 1+ \frac{x}{q} \right)  ^{q} $$
3416
Can you help me to prove that $$(x+y)^n\leq 2^{n-1}(x^n+y^n)$$ for $n\ge1$ and $x,y\ge0$.  I tried by induction, but I didn't get a result.
84323
The question reads     $$U_n = (1+x)^n - 1 - nx$$   Show that $U_2 \geq 0$   Hence or otherwise show that $(1+x)^n \geq 1 + nx$ for all $x \gt -1$.   Obviously the $U_2 \geq 0$ is very easy, I can do that without any trouble but I cannot see how it links to the 2nd part of the question.
13042
Trying to prove that above stated question for $n \geq 1$. A hint given is that you should use $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$. Using this, I think I reduced it to $\frac{1}{2}(\frac{1}{n^2}-(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}))$. Just not sure if it's correct, and what to do with the second half.
94859
Let $x = (x_1, x_2,...,x_n)\in\{0,1\}^n$ for $n\in\mathbb{N}$. Prove that if one-way functions (OWFs) exist, then there exists a one-way function $f$ such that for every bit $i\in[1,n]$ there exists an algorithm $A_i$ such that: $$\Pr[A_i(f(x)) = x_i] \geq \frac{1}{2} + \frac{1}{2n}$$ I'm having trouble understanding this.