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11

Hint: Note that $1+ab>0$ and $1+ab-a-b=(1-a)(1-b)$


6

First let $$ x = \sqrt{\frac{a}{a+8}}, \,\, y = \sqrt{\frac{b}{b+8}}, \,\, z = \sqrt{\frac{c}{c+8}} \,\, $$ Then $1 > x,y,z > 0$ and $$ a = \frac{8x^2}{1 - x^2}, \,\, b = \frac{8y^2}{1 - y^2}, \,\, c = \frac{8z^2}{1 - z^2},\,\, $$ So the question transforms to this: Given that $1 > x,y,z > 0, \, \, \frac{512x^2y^2z^2}{(1 - x^2)(1 - y^2)(1 ...


4

Let $f(x)=(1+4\sqrt x+x)\log x-(6x-6)$. Then, we have $$f'(x)=\left(\frac{2}{\sqrt x}+1\right)\log x+\frac{1+4\sqrt x+x}{x}-6$$ $$f''(x)=\frac{x-\sqrt x\ \log x-1}{x^2}$$ Let $g(x)=x-\sqrt x\ \log x-1$. Then, we have $$g'(x)=\frac{-2-\log x+2\sqrt x}{2\sqrt x},\quad g''(x)=\frac{\log x}{4x^{3/2}}$$ Now note that $g''(x)\gt 0$ for $x\gt 1$ and that ...


4

$$\int_{0}^{1}|P(x)|\,dx \leq \sum_{i=0}^{n}\int_{0}^{1}|a_i|x^i\,dx = \sum_{i=0}^{n}\frac{|a_i|}{i+1}\color{red}{\leq}\sqrt{\sum_{k=1}^{n+1}\frac{1}{k^2}}\leq\sqrt{\zeta(2)}=\frac{\pi}{\sqrt{6}}$$ where $\color{red}{\leq}$ follows from the Cauchy-Schwarz inequality.


4

Set $y=1-x$ and solve $ny^{n-2}=\frac{1}{5}\,$ for $n.$ A solution in terms of LambertW is $$n = \frac{W_0(\frac{1}{5}y^2\ln y)}{\ln y}.$$ Example $x=0.75$, then $n > 0.01272241834262.$ Check: with $n_1=0.0127\,$ you get $n_1(1-x)^{n_1-2} \approx 0.19965\,$ and with $n_2=0.0128\,$ the numbers are $n_2(1-x)^{n_2-2} \approx 0.20120$ Edit: The other real ...


4

If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq 1$, then $a,b,c,d>1$ and $$\begin{align} 0\leq abcd-bcd-cda-dab-abc=&(1-a)(1-b)(1-c)(1-d)-1+(a+b+c+d) \\&-(ab+ac+ad+bc+bd+cd) \\ =&\frac{9}{16}-1-a(b-1)-b(c-1)-c(d-1)-d(a-1) \\ &-ac-bd \\ <&-\frac{7}{16}\,, \end{align} $$ which is a contradiction. Hence, ...


4

Just consider e.g. $n=2$ and $x_1 = x_2 = \frac{1}{4}$. Note that the AM-GM inequality is of a very similar form; it states that $$x_1+\ldots+x_n \geq n (x_1 \cdots x_n)^{\frac{1}{n}}.$$ This means in particular that the inequality from the OP holds true whenever $x_1+\ldots+x_n \geq 1$.


4

Almost everything is perfect, and you can show a chain of inequalities in steps. The only detail is that you also need $a\ne b$ to claim $0<(\sqrt a - \sqrt b)^2$ (Which you have, you just didn't cite), since otherwise the difference could be 0.


3

By Holder's Inequality, $$S^2\sum_{cyc} a^2b^2\ge (a^2+b^2+c^2+d^2)^3$$ So it remains to note that by AM-GM $$\sum_{cyc}a^2b^2 = (a^2+c^2)(b^2+d^2) \le \frac{\left((a^2+c^2)+(b^2+d^2)\right)^2}4$$


3

I assume $E$ is convex and $H$ is a real Hilbert space. Wlog, $y^\ast = 0$. Since $E$ is convex and $0$ is the closest point in it to $x$, we have $\langle y,x\rangle \le 0$ for all $y\in E$. If $\langle x,x'\rangle \le 0$ then $$ \|x'-x\|^2 + \|x'-y'\|^2 \ge \|x'-x\|^2 = \|x'\|^2 + \|x\|^2 - 2\langle x,x'\rangle \ge \|x'\|^2 $$ as desired. If $\langle ...


3

Take \begin{align} H &= \mathbb{C} \\ E &= \lbrace 0, 4 \rbrace \\ x &= 1 \\ x' &= 3 \\ y' &= 4 \end{align} Then $y^* = 0$, and the conjecture is proven false.


3

First of all, there is really no reason to have the $1-x$, you can do everything again with $n y^n$, since $f(x)=1-x$ is a bijection from $(0,1)$ to itself. Now to have $y^n < \delta$, it is enough to have $n>\frac{\ln(\delta)}{\ln(y)}$. You need $\delta=\varepsilon/n$. So you in principle need $n>\frac{\ln(\varepsilon) - \ln(n)}{\ln(y)}$. The ...


2

For $m,n\in\mathbb{N}$ with $m\leq n$, let $S_{m,n}:=\prod_{i=m}^n\,\frac{2i-1}{2i}$. Then, $$S_{m,n}^2=\prod_{i=m}^n\,\left(\frac{2i-1}{2i}\right)^2 \leq \prod_{i=m}^n\,\left(\frac{2i-1}{2i}\right)\left(\frac{2i}{2i+1}\right)=\prod_{i=2m-1}^{2n}\,\frac{i}{i+1}=\frac{2m-1}{2n+1}\,.$$ Therefore, $S_{m,n}\leq\sqrt{\frac{2m-1}{2n+1}}$. In particular, ...


2

I don't know how to use Maple for this. I'll explain how to do this kind of thing by hand, the way I answered your previous question of this type. What we want is to rewrite the given expression in terms of sums of products of things whose signs we know, like $a-b-x$ and $b+y-c$. So let's look at the first term: $$ a^3 b $$ This has an $a$ in it, so maybe ...


2

Consider the telescoping sum $$a_1\cdots a_k-b_1\cdots b_k=\sum_{i=1}^k a_1\cdots a_{i-1}(a_i-b_i)b_{i+1}\cdots b_k. $$


2

As I said in my comment above, one can define: $\alpha=1/BC$, $\beta=1/AC$, $\gamma=1/AB$ so that we are required to find the point $P$ such that $\alpha PA + \beta PB + \gamma PC$ attains its minimum value. Following the method outlined here let's denote by $\vec{i}$, $\vec{j}$ and $\vec{k}$ the unit vectors along $\vec{PA}$, $\vec{PB}$ and $\vec{PC}$ ...


2

By the definition of $T$ we have $Z_T>2\alpha$ whenever $T<\infty$. If, in addition, $|Z_n-Z_T|\leq \alpha$ we get $$|Z_n|-\alpha=\alpha + |Z_n|-2\alpha > \alpha + |Z_n|-|Z_T| \geq \alpha - |Z_n-Z_T|\geq 0.$$


2

Here is another rather elementary answer. Let $a_m=\frac{1}{4m}\binom{2m}{m}$. We show the following is valid \begin{align*} \frac{1}{2\sqrt{m}}\leq a_m \leq \frac{1}{\sqrt{2m+1}}\qquad\qquad m\geq 1\tag{1} \end{align*} We start similarly to @CuriousGuest \begin{align*} ...


2

Presumably $A$ and $B$ are real symmetric. We have \begin{align} \operatorname{tr}\left(A(B+A)^{-2}\right) &=\operatorname{tr}\left((B+A)^{-1}A(B+A)^{-1}\right)\\ &\le\operatorname{tr}\left((B+A)^{-1}(B+A)(B+A)^{-1}\right)\\ &=\operatorname{tr}\left((B+A)^{-1}\right)\\ &\le\operatorname{tr}\left(A^{-1}\right). \end{align}


2

hint: $1-\dfrac{2a}{2a+b}=\dfrac{b}{2a+b}=\dfrac{b^2}{2ab+b^2}$


2

Hint: $$1_{\{|X|>a\}} \leq \left| \frac{X}{a} \right|^4 1_{\{|X|>a\}} \leq \left| \frac{X}{a} \right|^4. $$ Now integrate both sides.


2

You almost got it, the solution is to set $t_1=x^4 , t_2=y^4, t_3 = z^4, t_4=16$. This gives you $$ 2xyz \leq \frac{x^4+y^4+z^4+16}4 $$, which is what you want.


2

This is equivalent to $$b^{kb}-b^{ka}\ge a^{kb}-a^{ka}$$ Dividing, note that the RHS is negative ($\alpha^{x}$ is decreasing for $\alpha \in (0,1)$), so we want to show $$\frac{b^{kb}-b^{ka}}{a^{kb}-a^{ka}}\le 1$$ Let $f(x)=b^{kx}$ and $g(x)=a^{kx}$. Then by Cauchy's Mean Value Theorem, there is an $c \in (a,b)$ with ...


2

If there is a prime greater than $n$ and at most $nm-2$, it divides the left side but not the right side. Bertrand's hypothesis almost gives this to you.


1

Hint: There is a theorem that says that there is always a prime between $n$ and $2n$ (and actually, the constant $2$ can be reduced if a lower bound on $n$ is used). So if $m > 2$, you should be able to find a prime strictly in between $n$ and $nm - 2$, at least if $n > 3$, and then your equation can't hold. Also, using the lowered constant, (for $n$ ...


1

Put $x_1=x_2=\cdots=x_n=x$. The inequality becomes $(nx)^{n-1}\ge1$, which is not generally true.


1

I haven't had the time to check all the details to the end, but I think that a solution follows from the observation that $p_n \leq \frac{1}{2}(p_{n-1}+p_{n+1})$. Putting $a_n = \frac{p_n}{p_{n-1}}$ yields $a_n \geq 2-\frac{1}{a_{n-1}}$ and you can use this to prove by induction that $$a_{n+1}\geq \frac{\sqrt{2n}+1}{\sqrt{2(n+1)}},$$ as noted in the ...


1

The Maple code sol := solve({(1/2)*a*t^2-b*t+1 = 0, (1/6)*a*t^3-(1/2)*b*t^2+t-l = 0, (1/2)*c*t^2-d*t+d-(1/2)*c-1 = 0, (1/6)*c*t^3-(1/2)*d*t^2+(d-(1/2)*c-1)*t+(1/3)*c-(1/2)*d+1-l = 0, a*t-c*t-b+d = 0}, {a, b, c, d, t}, parametric = full): allvalues(sol); does the job. See its huge nested output in this Maple 2015 file. The one can be viewed by free ...


1

Puting $t=0$ in Eq(1) and $t=1$ in Eq(3) shows us that both $0$ and $1$ are not possible values for t. So, dividing by $t$ and $t-1$ should be fine. Solving Eq(1) and Eq(2) for $a$ and $b$ in terms of $l$ and $t$ we get \begin{eqnarray*} a &=& \frac{6(t-2l)}{t^3} \\ b &=& \frac{2(2t-3l)}{t^2} \\ at - b &=& \frac{6t-12l-4t+6l}{t^2} = ...


1

Geometric Proof: In this proof, $M$ is not assumed to be inside $ABC$. Let $P$ and $Q$ be the points such that $APBC$ and $MQBC$ are parallelograms. Note that $APQM$ is also a parallelogram. Hence, $AP=BC=MQ$, $BP=CA$, $BQ=MC$, and $PQ=MA$. Consider the quadrilateral $APQB$. We have by Ptolemy's Inequality that $$MC\cdot BC+MA\cdot AB=AP\cdot BQ+PQ\cdot ...



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