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11

Substituting $t=|\frac {x-3}{x+1}|$ will help. Can you see it?


8

Note that $$ \frac{1}{i+1}+\frac{1}{k-i+1}=\frac{k+2}{(i+1)(k-i+1)} $$ So, (because $(a+b)^2\leq 2a^2+2b^2$): $$ \frac{(k+2)^2}{(i+1)^2(k-i+1)^2}\leq\frac{2}{(i+1)^2}+\frac{2}{(k-i+1)^2} $$ Adding we get $$ \sum_{i=0}^{k}\frac{(k+2)^2}{(i+1)^2(k-i+1)^2} \leq 4\sum_{i=0}^{k}\frac{1}{(1+i)^2}\leq 4\zeta(2)=\frac{2\pi^2}{3} $$ So $$ ...


6

Using the substitution $a=\frac xy$, $b=\frac yz$ and $c=\frac zx$, we want to prove that $$ \sum_{cyc}\frac {xy}{z^2}\geq \frac xy+\frac yz+\frac zx $$ where we are taking cyclic sums. Using AM-GM, we find that $$ \frac{\frac{xy}{z^2}+\frac{xz}{y^2} +\frac{xz}{y^2}}3 \geq \sqrt[3]{\frac{x^3}{y^3}}=\frac xy $$ Applying this three times yields the required ...


5

For any $C$, one can always find an $X$ such that $\det(X+iC)=0$ : if $C$ is not invertible, simply take $X=0$. If $C$ is invertible, take $X=CA$ where $A$ is any real matrix having a two-dimensional invariant subspace ${\sf span}(v_1,v_2)$ with $Av_1=v_2$, $Av_2=-v_1$ (so that $\pm i$ are eigenvalues of $A$). Then $$ ...


4

Define the following two vectors \begin{align} \textbf{a}&=(a_1,a_2,...,a_n),\\ \textbf{b}&=(b_1,b_2,...,b_n) \end{align} the dot product is defined such that $|\textbf{a}\cdot \textbf{b}|^2 = |\textbf{a}|^2|\textbf{b}|^2\cos^2\theta$ where $\theta$ is the angle between the two vectors $\textbf{a}$ and $\textbf{b}$. So the desired condition holds if ...


4

Observe that $\frac{1}{n}$ is decreasing $(1)$. $$\int\limits_{1}^{n}{\dfrac{1}{x} \, dx}=\int\limits_{1}^{2}{\dfrac{1}{x}\, dx}+\int\limits_{2}^{3}{\dfrac{1}{x}\, dx}+\cdots +\int\limits_{n-1}^{n}{\dfrac{1}{x}\, dx} \tag{2}$$ For an integrable function $f$, if $ m \le f(x)\le M \quad \forall x\in [a,b]$ then $$m(b-a)\le\int\limits_{a}^{b}{f(x) dx} \le ...


4

Hint (for a start): $$t = \frac{x-3}{x+1}$$


4

Note that $$ (n!)^2 = \prod_{i=1}^n i(n+1-i) \ge \prod_{i=1}^n n = n^n $$ Hence $n! \ge n^{n/2}$. This gives $$ (2n)! \ge (2n)^n = 2^n n^n $$ For $n \ge 2$ we have $2^n \ge n^2$, hence $(2n)!\ge n^{n+2}$.


4

Hint: Consider $f(x)=\frac{\sqrt 2}{2}\sin x- e^x+1$. Note that $f(0)=0$. Now Find $f'(x)$. Then conclude.


3

If $(k-1)<\log_2n\le k$, i.e. $2^{k-1}< n\le 2^k$, then $a_{2^{k-1}}\le a_n\le a_{2^k}\le ka_2$ (by nondecreasing and your result). We conclude that $$ a_n\le a_2\cdot \lceil \log_2 n\rceil \le a_2(\log_2n+1)\le 2a_2\log_2n$$ where the last step uses $\log_2 n\ge1$ (from $n\ge 2$) and $a_2\ge 0$.


3

$$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 \Rightarrow \\ \left(\left|\frac{x-3}{x+1}\right|\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 $$ $$\Delta=\left(-7\right)^2-4 \cdot 10=49-40=9$$ $$\left|\frac{x-3}{x+1}\right|_{1,2}=\frac{-\left(-7\right) \pm \sqrt{\Delta}}{2}=\frac{7 \pm \sqrt{9}}{2}=\frac{7 \pm 3}{2}$$ ...


3

Step 1. Prove that $$ x^2+yz+x+1\geq x(x+y+z+1) $$ by using the expansion of $(x-y-z)^2\geq 0$ and $x^2+y^2+z^2=2$. Step 2. Observe that $$ x+y+z\leq\sqrt{2(x^2+(y+z)^2)}=2\sqrt{1+yz} $$ Step 3. By Step 1, we have $$ P\leq\frac{x^2}{x(x+y+z+1)}+\frac{y+z}{x+y+z+1}-\frac{1+yz}{9}=1-Q $$ with $$ Q=\frac{1}{x+y+z+1}+\frac{1+yz}{9}. $$ By step 2 we have $$ ...


3

The unit circle is parametrized by $\theta\mapsto(\cos\theta,\sin\theta)$: $\hskip 1.5in$ Consider an angle $0\le\theta\le\frac{\pi}{2}$, as depicted above. The red line's length is $\theta$, and $\sin\theta$ is the purple length. Since they both travel the same vertical distance but the red one also travels horizontal distance, the red line is longer, ...


2

http://en.wikipedia.org/wiki/Hoeffding%27s_inequality gives a way to bound the difference between $\frac{1}{\sqrt{3}}$ and the given integral. If we take $X_1,\ldots,X_n$ as indipendent random variables with a uniform distribution over $[0,1]$, then $X_1^2,\ldots,X_n^2$ are bounded indipendent random variables with density function $$ ...


2

The evaluation of limit is essentially same as here. We have, $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} \left(\dfrac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{n}\right)dx_{1}dx_{2}\cdots dx_{n}=\dfrac{1}{3}$ We can show $\displaystyle \lim\limits_{n\to\infty}\int_{0}^{1}\cdots\int_{0}^{1} ...


2

To be 100% vigorous with the proof, you'd need to show that $k^3 > 2k +1$ for $k>4$. And since it's an easy one, go ahead and do it with induction again. Otherwise, you're 100% spot on with your induction. Note that you don't need to use strong induction. Weak induction would suffice. It only needs to hold for $k$ into order to prove $k+1$. But that ...


2

When $t \equiv \frac{x-3}{x+1} > 0$, the solution to $ t^2 - 7t + 10 <0$ is $ 2<t<5$. (Factor the quadratic.) When $t < 0$, the solution to $ t^2 + 7t + 10 <0$ is $ -2<t<-5$. On the $t < 0$ branch, we need to solve $-2 < \frac{x-3}{x+1} < 5$. We break this up into two possibilities, $x < -1$ and $x > -1$, because ...


2

To prove the first inequality it suffices to prove that $(x+y)^{p}\le x^{p}+y^{p}$ for $x,y\ge0$ and $p\in(0,1)$. Notice that we can use this as follows: $\vert x\rvert^{p}\le\lvert x-y\rvert^{p}+\lvert y\rvert^{p}$ then reversing the roles of $x$ and $y$. Notice that this Lemma can be proven by reducing to the case when $y=1$ then looking at ...


2

Hint: 1) From $(\frac{a+b}{2})^2\le\frac{a^2+b^2}{2}$ verify that $(k+1)^2 < 2[(i+1)^2+(k-i+1)^2]$, 2) $\sum_{n\ge 1}\frac{1}{n^2} = \frac{\pi^2}{6}$.


2

One can forget $p(s)$, $r(s)$ and the rest and simply try to show that, for every $a\lt b$, $$\|u\|_a\leqslant\|u\|_b.$$ To wit, considering $v=|u|^a$ and $p=b/a\gt1$, note that Hölder inequality yields $$ \int |u|^a=\int v\leqslant\left(\int v^p\right)^{1/p}=\left(\int |u|^b\right)^{a/b}, $$ that is, $$ \left(\int |u|^a\right)^{1/a}\leqslant\left(\int ...


2

First, note that next recurrent relation is true for sequence $(a_n)$: $$ a_0=1,\\ a_{n+1} = a_n+\frac{1}{a_n}, \qquad n\ge 0.\tag{1} $$ Yes, $a_0=1$, $a_1 = 1+\frac{1}{1}=2$ for both definitions; if $a_{n+1}$ is defined by $(1)$, then $$ a_{n+1} = a_n+\dfrac{1}{a_n} = a_n + \dfrac{1}{a_{n-1}+\frac{1}{a_{n+1}}} = a_n + \dfrac{a_{n-1}}{1+a_{n-1}^2}, \qquad ...


2

A proof just with Cauchy-Schwarz: From Cauchy-Schwarz, one have that $$ (\sum_i \frac{1}{x_i})(\sum_i \frac{x_i}{i^2}) \ge (\sum_i \frac{1}{i})^2. $$ But since $\sum_i \frac{1}{x_i} \le \sum_i \frac{1}{i}$, one have that $$ \sum_i \frac{x_i}{i^2} \ge \frac{(\sum_i \frac{1}{i})^2}{\sum_i \frac{1}{x_i}} \ge \frac{\sum_i \frac{1}{i}}{\sum_i \frac{1}{x_i}} ...


2

There is one step that is not (always) valid. The inequality $$x^2 + (k+1)xy \geqslant \frac{1}{x}(x^2+(k+1)xy)$$ only holds for $x+(k+1)y \geqslant 0$, if $x > 1$. If $x+(k+1)y < 0$ and $x > 1$, you have a strict inequality in the other direction, $$x^2 + (k+1)xy < \frac{1}{x}(x^2+(k+1)xy).$$ However, in that case, the desired inequality ...


2

First of all: you cannot take derivatives (and even if so, your middle one is incorrect) to preserve inequalities. That said, let's go over it: 1) It is enough to show that $\sin x\le x$ when $x\ge 0$ because $\sin x$ is an odd function, so for $x<0$ write $-y=x$ whence the first inequality implies $$-\sin -y \le y\iff \sin(-y)\ge -y\iff \sin x\ge x$$ ...


2

For any $k\geq 1$ we have: $$(2^{k+3}-1)(2^{k+2}-1)=2^{2k+5}-3\cdot 2^{k+2}+1\leq16(2^{2k+1}-1)$$ and for any $k\geq 3$ we have: $$3(2k+1)\geq 16,$$ hence we have to check only the cases $k=1,2$.


2

I think you are doing a mistake here. $ \sin(x)+\cos(y)+\tan^2(y)+\cot^2(x)+5 = (\sin(x)+ \csc^2(x)) + (\cos(y)+ \sec^2(y))+ 3$ . Not, what you have written. Considering that to be a typo, Notice that the things in the bracket are independent, so we just find minimum value of each separately. Consider $f(x)=\sin(x)+\csc^2(x) \ge \sin(x)+\csc(x) \ge 2$. ...


2

Taken literally, property (2) does not imply (1). Indeed, let $f(x)=-1$ if $|x|=1$ and $f(x)=0$ otherwise. Then (2) holds but (1) fails for $x=0$ and $r=1$. But generally, one considers locally integrable functions up to equivalence a.e. And property (2) implies that $f$ agrees a.e. with a subharmonic function, for which (1) holds by definition. Here are ...


1

By the Rearrangement Inequality (but we don't need anything that general) the left side is minimized, for fixed $x_i$, if the $x_i$ are increasing. And then the minimum is reached if the $x_i$ are as small as possible, which gives $x_i=i$. Remark: If we don't want to quote the Rearrangement Inequality, it is clear that if $i\lt j$ and $x_i \gt x_j$, then ...


1

first we check $abc=0$,if $a=0 ,\implies 2\sqrt{bc}\le b+c$,it is true.the "=" will hold when $b=c$ in case $abc \not=0$ let $c=Max${$a,b,c$}, we discuss two cases: $a^2+b^2+c^2 \ge 2(ab+bc+ac) \implies c\ge a+b+2\sqrt{ab} \implies ab \le \dfrac{c^2}{16}$ LHS $\le\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+\dfrac{c^2}{16}}\iff ...


1

I propose that the question has a typo and that the inequalities in the index of the summation cannot be strictly less than and need to be less than or equal to. Here is why. Consider the case where $n=2$ and $r=3$. Clearly, the left hand side is an application of binomial theorem and we have $a_1^3+3a_1^2a_2+3a_1a_2^2+a_2^3$ The right hand side is (from ...



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