Suppose we have a positive sequence $\{x_i\}$ where $i=1, 2, \dots , n$. And let $m=\mathrm{min}(x_i), \; M = \mathrm{max}(x_i)$. Notice that $m, M&gt;0$. Prove this inequality (discrete version): $$ \frac{\sum_{i=1}^{n} x_i }{n} \cdot \frac{ \sum_{i=1}^{n} \frac{1}{x_i} }{n} \le \frac{m+M}{2} \cdot \frac{ \frac{1}{m} + \frac{1}{M} }{2} \;.$$ Or you can prove this continuous version, where $f(x)$ is positive and continuous on $[0, 1]$ , and $m$ and $M$ are min and max values of $f(x)$ on $[0, 1]$, respectively: $$ \int_0^1 f(x) \,\mathrm{d}x \int_0^1 \frac{1}{f(x)} \,\mathrm{d}x \le \frac{m+M}{2} \cdot \frac{ \frac{1}{m} + \frac{1}{M} }{2} \;.$$ Important: I do have a proof, because the continuous version is actually an excercise of a workbook of mine. But that proof is too tricky, lacking insight as to why the product of two types of &quot;average&quot;s is controlled by their max and min values. So I brought it up here in seeking of an insightful solution. Hopefully finding a proof directly to the discrete version, for it is easier to imagine with my brain.
124068
Let $f,g$ be two continuous positive functions over $[a,b]$ Let $m_1$ and $M_1$ be the minimum and maximum of $f$ Let $m_2$ and $M_2$ be the minimum and maximum of $g$ Prove that $$\sqrt{\int_a^bf^2 \int_a^b g^2}\leq \frac{1}{2}\left(\sqrt{\frac{M_1M_2}{m_1m_2}}+\sqrt{\frac{m_1m_2}{M_1M_2}}\right)\int_a^bfg$$  I can't make any significant progress on this one... Thanks for any hint.
40499
If $a, b, c$ belong to positive real numbers then prove that  $$\frac{2}{b+c} +\frac{2}{c+a}+\frac{2}{a+b} \le \frac 1a +\frac 1b +\frac 1c.$$  I applied AM-GM inequalities on both sides but I cant even figure out what to do on left side.
18290
Problem Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent. One Proof This proof is based on the following inequality  $$\frac{1}{n+1}&lt;\ln \left(1+\frac{1}{n}\right)&lt;\frac{1}{n}$$  where $n=1,2,\cdots$, which will be used repeatedly. On one hand, we obtain that $$\ln 2-\ln 1&lt;1,~~\ln 3-\ln 2&lt;\frac{1}{2},~~\ln 4-\ln 3&lt;\frac{1}{3},~~\cdots,~~\ln (n+1)-\ln n&lt;\frac{1}{n}.$$ Adding up all of these，we have that $\ln(n+1)&lt;1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}.$ Hence,$$x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)&gt;\frac{1}{n+1}&gt;0.$$ This shows that $x_n$ is bounded below. On the other hand,$$x_n-x_{n+1}=-\frac{1}{n+1}+\ln(n+1)-\ln n=\ln \left(1+\frac{1}{n}\right)-\frac{1}{n+1}&gt;0.$$ This shows that $x_n$ is decreasing. Combining the two aspects, according to Monotone Bounded Theorem, we can assert that $\lim\limits_{n \to \infty}x_n$ exists. Let $\gamma$ (so-called Euler–Mascheroni Constant) denote the limit, i.e. $$\gamma=\lim_{n \to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n\right),$$which equals $0.577216 \cdots$. We may also express that as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\gamma+\ln n+\varepsilon_n,$$where $\varepsilon_n$ represents an infinitesimal related to $n$ under the process $n \to \infty$.
11593
So I'm trying to prove the following two inequalities: $$\frac{\sqrt{n}}{2} \leq \phi(n) \leq n.$$ The upper bound we get from simply noting that $\phi(n) = n \prod_{p | n}\left( 1 - \frac{1}{p}\right)$ and the fact that $(1 - \frac{1}{p}) \leq 1$. But how can we get the lower bound? I tried using the same expression for $\phi(n)$ but it seems to not really give me the inequality. Can you help?
70462
I want to show that $$\frac{x}{1+x}&lt;\log(1+x)&lt;x$$ for all $x&gt;0$ using the mean value theorem. I tried to prove the two inequalities separately.  $$\frac{x}{1+x}&lt;\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) &lt;0$$  Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}&lt;0$$ for all $x &gt; 0$,  $f(x)&lt;0$ for all $x&gt;0$.  Is this correct so far?  I go on with the second part: Let $f(x) = \log(x+1)$. Choose $a=0$ and $x&gt;0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with   $f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.  Since $$x_0&gt;0 \Rightarrow  \frac{1}{x_0+1}&lt;1.$$ $$\Rightarrow 1 &gt; \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x&gt; \log(x+1)$$
48085
I want to prove that for $a, b, c &gt; 0$ we have $\frac{a}{2a + b} + \frac{b}{2b + c} + \frac{c}{2c + a} \leq 1$.  My approach:  I know that each of the individual terms is lesser than $\frac{1}{2}$ because of it's form. I am familiar with the Cauchy-schwarz inequality and the AM-GM-HM inequality. I tried using AM-GM but could not get anywhere because of the way the inequality is structured. Similarly I tried using Cauchy-schwarz as well.  I just need some intuition/hints on how to actually reduce this to a feasible form to solve and not the actual answer because it does not provide me with the necessary intuition which I need as I am a beginner when it comes inequalities. If I can't use these 2 inequalities, is this problem best tackled with some other inequality?
63372
Let $a_1,a_2,\ldots ,a_n$, be positive real numbers and  S = $a_1+a_2+a_3+\cdots+a_n$. Use the Cauchy-Schwarz inequality to prove that   $$\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}$$ and $$\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)$$  Hint: Apply the inequality to:   $\sum \frac{S}{a_i}$ and $\sum \frac{a_i}{S}$ for one case,     and $\sum a_i(S-a_i)$ and $\sum \frac{a_i}{S-a_i}$ for the other case.
55163
Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d = 4$. Prove that $$\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a .$$   How can I approach this using only the ? Are there any other methods that do not involve the AM-GM inequality?
48460
Let $x_1,x_2,\ldots,x_n &gt; 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).$$  This is Exercice 1.48 in the book &quot;Inequalities-A mathematical Olympiad approach&quot;. Attempt I tried using HM-GM and I got $\left ( \dfrac{1}{x_1x_2\cdots x_n}\right)^{\frac{1}{2n}} \geq \dfrac{n}{\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}}} \implies \dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_1}}+\cdots+\dfrac{1}{\sqrt{x_n}} \geq n(x_1x_2 \cdots x_n)^{\frac{1}{2n}}$. But I get stuck here and don't know if this even helps.
97356
I'm having some difficulty solving this problem :     Given the sequence $(a_n)$ with $a_n \geq a_{n+1} \geq 0$ show that if $$\sum \sqrt{a_n a_{n+1}} &lt; \infty \implies \sum a_n &lt; \infty.$$    I first thought of using the Arithmetic Mean-Geometric Mean Inequality:   $$\forall{a,b} &gt; 0: \frac{a+b}{2} \geq \sqrt{ab}.$$  Therefore, since $a_n \geq a_{n+1} \geq 0$ we have: $$\frac{a_n+a_{n+1}}{2} \geq \sqrt{a_n a_{n+1}} \;\; \forall{n} \in \mathbb{N}.$$  but I can't use the Direct Comparaison Test here since it will only work to show the other side of the implication $\sum a_n &lt; \infty \implies \sum \sqrt{a_n a_{n+1}} &lt; \infty$.
104836
I have to prove the following inequality:     $$\forall n,m \in \mathbb{N},i\in\mathbb{N_0}:n&lt;m \implies \frac1{n^i}\binom{n}i \leq\frac1{m^i}\binom{m}i $$   I used this fact to prove an other inequality, without proving this one first, but now I am stuck. Is induction the way to do this?
48112
This question came from the prelim exam I took last month. I have a proof that seems a bit unwieldy to me (posted as an answer), so I'm opening it up to ask if there are other ways of showing this.     Let $x$ be any positive real number, and define a sequence $\{a_n\}$ by   $$ a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} $$   where $[x]$ is the largest integer less than or equal to $x$. Prove that $\displaystyle{\lim_{n \to \infty} a_n = x/2}$.
65486
I would like to verify my proof of the following:  Let $A=\{\frac{1}{n}-\frac{1}{m}:m, n \in \mathbb{N}\}$.  I want to show that $-1$ and $1$ are the infimum and supremum respectively.  First I will show that $-1$ is the infimum.  To show this I will first demonstrate that it is indeed a lower bound.  Observe: $$\frac{1}{n}-\frac{1}{m} \geq \frac{1}{n}-1 \geq -1.$$  I now claim that $-1$ is the greatest lower bound.  So, by the archimedean property, there exists an $x \in \mathbb{R}$, $x&gt;0$ and $n' \in \mathbb{N}$ such that $1&lt;n'x$.  Thus, $\frac{1}{n'}&lt;x$ and $$\frac{1}{n'}-1&lt;x-1.$$  From the above it is clear that $\frac{1}{n'}-1 \in A$.  Also, it is clear that $-1&lt;-1+x$.   Let $y=\inf(A)$.  So we let $$-1&lt;y\leq-1+x.$$  By completeness there exists an $r\in\mathbb{Q}$ such that $$-1&lt;r&lt;y\leq-1+x.$$  Thus $y$ is not the infimum.  I was going to note that the supremum was 1 through the relationship between $$\inf(A)=-\sup(-A).$$
26150
Show that. for any discrete random variable X that takes on values in the range [0,1]. Var[X] $\le$  1/4.  I translate it into a inequality like this: $x_1, x_2, x_3 \cdots ,x_n$ where $0 \le x_i \le 1$, and $p_1, p_2, p_3 \cdots ,p_n$ where $p_1+ p_2+ p_3 \cdots +p_n = 1$, prove that $\sum _1^nx_i^2p_i - (\sum _1^n x_ip_i)^2 \le {1\over 4}$ , how to prove it?  At first I tried cauchy inequatlity, but I fail :(
46933
I tried to solve this problem but I could not continue  Let $f: [a, b​​]\to \mathbb{R}$ be a continuous function.  Prove that if $x_1, x_2,\ldots,x_n  \in (a,b​)$, then exists $x_0 \in (a, b​​)$ such that $$f (x_0) =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}$$  the function is continuous, so $\forall\ x_1, x_2,\ldots,x_n  \in(a, b​​)$ $$\begin{align*} \lim_{x \to \ x_1}f(x)&amp;=f (x_1)=L_1\\ \lim_{x \to \ x_2}f(x)&amp;=f (x_2)=L_2\\ \lim_{x \to \ x_3}f(x)&amp;=f (x_3)=L_3\\ &amp;\vdots\\ \lim_{x \to \ x_n}f(x)&amp;=f (x_n)=L_n\\ \end{align*}$$  We have that:  $$\lim_{x \to \ x_0}f(x)=f (x_0)=\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}= \frac{1}{n}\sum_{j=1}^n f(x_j)$$ Now $$\min\left \{ L_1,L_2,L_3,\cdots,L_n \right \}\le\frac{1}{n}\sum_{j=1}^n f(x_j)\le\max\left \{ L_1,L_2,L_3,\cdots,L_n \right \}$$  and at this point I'm lost ...
81903
Let $X_n$, $n=1,2,3,...$ be a sequence of independent (not necessarily identically distributed) random variables, let $S_n=\sum_{i=1}^nX_i$. Prove the following maximal inequality: For all $t&gt;0$,$$\mathbb{P}\left(\max_{1\leq i\leq n}|S_i|\geq t \right)\leq 3\max_{1\leq i\leq n}\mathbb{P}\left(|S_i|\geq\frac{t}{3} \right)$$
97993
I was asked to prove an inequality: For any $n$ positive numbers $\{a_i\}$ with $a_{1}a_{2}\cdots a_{n} = 1$ and $m \geq n-1$ be a non-negative integer, $a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$  After several attempts, I think I had to use induction on both $m$ and $n$. In particular, I fix an $n$ and claim that the inequality holds for every $m$, and use induction on $n$.  The case where $n = 2$ could be done by rewriting the inequality as $a_1^2 + a_2^2 \geq a_1 + a_2$ and assume  that $a_1 \leq 1 \leq a_2$.   In the inductive step, given $a_1^k + a_2^k \geq a_1 + a_2$ , by writing $a_1^{k+1} + a_2^{k+1} = (a_1^{k} + a_2^{k})(a_1 + a_2) - a_1^{k}a_2^{k} (a_1 + a_2)$, the result follows.  However, this method seems to fail for $n&gt;2$, as the right-hand side of the inequality becomes more complicated. Should I try not to prove it by induction?