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8

Such formulas for such numbers exist for all even $k$. The formula is not as magical as it seems. Consider $$ 10^{ni}-m\sum_{s=0}^{n-1}10^{si}\;. $$ with fixed $m,n$ for all $i$. There are $n$ stretches of $O(i)$ nines in this number. Taking it to the $k$-th power yields $$ \left(10^{ni}-m10^{(n-1)i}\right)^k+R\quad\text{with}\quad R\in ...


6

Solution 1.(Partial solution) $x^n$ is a convex function on $\mathbb{R}^+$ for $n\geq 1$, thus by Jensen: $$\left(\frac{a+b}{2}\right)^n\leq \frac{a^n+b^n}{2}$$ Solution 2.('Hidden' usage of condition) For $n=1$, true. By induction: $$\left(\frac{a+b}{2}\right)^n=\left(\frac{a+b}{2}\right)^{n-1}\left(\frac{a+b}{2}\right)\leq ...


6

No, for $$ f(t)=t, $$ we have $f(0)=0$ and $$ \int_0^t s^{-1}f(s)\,ds=t=f(t),\quad\forall t>0, $$ so (your constant is $1$, and you actually have equality) $$ f(t)\leq 1\int_0^t s^{-1}f(s)\,ds,\quad\forall t>0. $$


6

A simple approach would be to simply count the numbers of $x_i$ taking the values $-1$, $1$ and $2$. Denote these numbers by $a$, $b$ and $c$, respectively. Then we have \begin{eqnarray*} \sum_{i=1}^nx_i &=&-1\times a+1\times b+2\times c &=& 19,\\ \sum_{i=1}^nx_i^2 &=&\ \quad1\times a+1\times b+4\times c &=& 99. ...


5

For $t \in (n, n+1)$, $$\frac{1}{n +1} < \frac{1}{t} < \frac{1}{n}.$$ Integrate from $n$ to $n + 1$ to get the desired result. (well done improving your question)


5

When $n$ is a positive integer, the function $f(x) = x^n$ is convex on $(0, +\infty)$, thus by Jensen's inequality: $$f\left(\frac{a}{2} + \frac{b}{2}\right) \leq \frac{1}{2}f(a) + \frac{1}{2}f(b),$$ gives the desired inequality. As pointed out by Andrews, the $f$ defined above is convex on the whole line only when $n$ is even for which case Jensen's ...


5

This is not true in general. For an interval of clear counterexamples, consider that for $x\in(\frac32\pi,2\pi)$ we have $$ \frac{\sin x}{x} < 0 < \cos x$$ Update after the question was amended to specify $0<x<\pi$: The mean value theorem says that $\frac{\sin x}{x} = \sin'(\alpha)$ for some $\alpha\in(0,x)$. We have ...


4

Consider the function $f(x)=\sin(x)$. Fix a point $y$ between $0$ and $\pi$. By applying the mean value theorem to the points $x=0$ and $x=y$, you know that for some point $z$ between $0$ and $y$, $$ \cos(z)=f'(z)=\frac{f(y)-f(0)}{y-0}=\frac{\sin(y)}{y}. $$ Since $\cos(x)$ is a decreasing function on the interval $0$ to $\pi$ and $y>z$, it follows that ...


4

$$ 2\left(\sqrt{i}-\sqrt{i-1}\right) = \frac{2}{\sqrt{i}+\sqrt{i-1}}\geq \frac{1}{\sqrt{i}}\tag{1} $$ hence: $$ \sum_{i=1}^{n}\frac{1}{\sqrt{i}}\leq 2\sum_{i=1}^{n}\left(\sqrt{i}-\sqrt{i-1}\right)=2\sqrt{n}.\tag{2}$$


4

$(a-1)(b-1)>0$ as $a,b>1 \implies ab+1>a+b $, or $2(ab+1)>ab+1+a+b=(a+1)(b+1)$. Similarly $(c+1)(d+1)<2(cd+1)$. Now $ab,cd>1$ then by similar method $(ab+1)(cd+1)<2(abcd+1)$ hence $4(ab+1)(cd+1)<8(abcd+1) $ therefore $(a+1)(b+1)(c+1)(d+1)<8(abcd+1)$


4

If $n < t < n+1 \to \dfrac{1}{n+1} < \dfrac{1}{t} < \dfrac{1}{n} \to \dfrac{1}{n+1} = \displaystyle \int_{n}^{n+1}\dfrac{1}{n+1} dt < \displaystyle \int_{n}^{n+1} \dfrac{1}{t}dt<\displaystyle \int_{n}^{n+1}\dfrac{1}{n}dt=\dfrac{1}{n}$


4

We apply the Cauchy Schwarz Inequality to the vectors $x = \left({\sqrt {\dfrac{a-1}{a}} , \sqrt {\dfrac{b-1}{b}} , \sqrt {\dfrac{c-1}{c}}}\right) $ and $y = \left({\dfrac{1}{\sqrt{bc}},\dfrac{1}{\sqrt{ac}}, \dfrac{1}{\sqrt{ab}} }\right)$ in $\Bbb R^3$. Then, $x \cdot y \le \lVert x\rVert \lVert y\rVert$ yields, $$ \dfrac{\sqrt{a-1} + \sqrt{b-1} + ...


4

It is enough to consider: $$ f(x)= \sin^3(x)-x^3\cos^2(x) $$ and prove it is an increasing function over $I=\left(0,\frac{\pi}{2}\right)$ by studying: $$ g(x)=\frac{f'(x)}{\cos(x)} = -3x^2\cos(x)+2x^3\sin(x)+3\sin^2(x).$$ $g(x)> 0$ follows from $\frac{\sin x}{x}> 1-\frac{x^2}{6}$ and $\cos(x)< 1$ over $I$.


3

After using C-S we need to prove that \begin{align*} &\displaystyle\frac{9}{6-2\cos\gamma\cos\left(\alpha-\beta\right)+1-2\cos^2\gamma}\geq\frac{6}{5} \\ \iff&\displaystyle4\cos^2\gamma+4\cos\gamma\cos\left(\alpha-\beta\right)+1\geq0 \\ ...


3

\begin{align} 1^2+2^2+...+&(n-1)^2 \quad\quad= \frac{n(n-1)(2n)}{6} = \frac{n^2(n-1)}{3} &< \frac{n^3}{3} \\ 1^2+2^2+...+&(n-1)^2+n^2 =\frac{n(n+1)(2n+1)}{6} =\frac{2n^3+3n^2+n}{6} =\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} &>\frac{n^3}{3} \end{align} Cleary our inequality is proven.


3

Then , $k(y\log x+x\log y)=k(xy^2+xyz-x^2y+x^2y+xyz-xy^2)=2kxyz$ Again , $k(z\log y+y\log z)=k(xyz+z^2y-zy^2+y^2z+xyz-z^2y)=2kxyz$. So, $k(y\log x+x\log y)=k(z\log y+y\log z)\implies x^yy^x=y^zz^y$ , as $k\not=0$. Similarly you can prove the second equality..


3

Another proof: For $0<x<1$, we have $x^m-\frac{x^{m+1}}{m} = x^m(1-\frac{x}{m}) >0$, thus $$e^{-x} = 1 - x + \frac{x^2}{2}-\frac{x^3}{6}+\cdots > 1 - x$$ by Taylor series. For $x\geq1$, we have positive $\geq$ non-positive. For $x<0$, take $t=-x$ and apply the usual $e^t\geq1+t$, easliy provable by Taylor series.


3

I think there is a typo as I read that book and own one and some typos were spotted by me as well. It should be: $a^2+\sqrt{a}+\sqrt{a} \geq 3\sqrt[3]{a^2\sqrt{a}\sqrt{a}} = 3a$


3

How about the following solution?


3

$$a+b+c+d+e+f=1\implies (a+c+e)(b+d+f)\le \frac14 \\ \implies (ab+bc+cd+de+ef)+(ad+af+cf+be)\le \frac14$$ We can show the first bracket can attain full value by say $a=b=\frac12$.


3

This looks suspiciously like one of the questions in my differential equations workshops this year. Anyway, let $z=||x-y||^2.$ Then $\frac{d}{dt}z=\frac{d}{dt}||x-y||^2$ which is equal to $$ =2 \begin{pmatrix} x_1-y_1 \\ x_2-y_2 \end{pmatrix} \begin{pmatrix} -x_1+2x_2+y_1-2y_2\\ -2x_1-x_2+2y_1+y_2 \end{pmatrix} \\ ...


3

Certainly not: if $\lvert f(x)\rvert<1$, on the contrary $\lvert f(x)\rvert>\lvert f^2(x)\rvert$. Same answer for any power $>1$. (I supposed you meant the square of the function, not the function composed with itself). It's the inverse situation for roots: if $\lvert f(x)\rvert<1$,$\lvert \sqrt{\lvert f(x)\rvert}>\lvert f(x)\rvert$.


3

Take $\mathbb R$ over $\mathbb R$ endowed with the absolute value norm, $v = 1$, and $v_0 = 2$. Then $$1 = |v| = |v - v_0 + v_0| \neq |v - v_0| + |v_0| = 1 + 2 = 3.$$ For inner product spaces in particular, $||x||^2 = \langle x, x \rangle.$ We have \begin{align*} ||x + y||^2 &= \langle x+y, x+y \rangle \\ &= \langle x,x \rangle + 2\langle x,y ...


3

Just consider e.g. $$(\Omega,\mathcal{F},\mu) := ((1,\infty),\mathcal{B}((1,\infty)),\lambda)$$ where $\lambda$ denotes the Lebesgue measure (restricted to $(1,\infty)$) and $$\phi(x) := x^2 \qquad g(x) := \frac{1}{x}.$$


3

Use the following: If $a_1, \ldots, a_n$ are distinct positive integers, then we can sort them according to their size $a_{n_1} \leq a_{n_2} \leq \ldots \leq a_{n_n}$ and get $$k \leq a_{n_k} ~ \forall k = 1, \ldots, n.$$ Hence, using Cauchy Schwarz, we compute \begin{align*} \left( \sum_{k=1}^n \frac{1}{k} \right)^2 & = \left( \sum_{k=1}^n ...


3

I make a sketch of the work here by Lagrange Multiplier. I am sure there are other more "elegant" techniques from exotic inequalities like AM-GM, Schur, CS, etc...I would be more than happy to see a proof of such work.Meanwhile, we can take this path... We are led to: $(2-x)x=(2-y)y=(2-z)z = \sqrt{3\lambda}\to \text { WLOG } x = y, x+z = 2$ is the worse case ...


3

A much easier (than my original) answer Let $c>0$ and let $$p_c(x)=(c+x)^n +(c-x)^n=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}c^{n-2k}x^{2k}$$ Note that $p_c(x)$ has only positive coefficients and every term has even power. So $p_c(0)\leq p_c(x)$ for all $x$. Then let $c=\frac{a+b}{2}$ and $x=\frac{a-b}{2}$. Then $c+x=a$ and $c-x=b$, and we get: ...


2

The cases $n=1,2$ are trivial, so lets suppose $n\ge 3$. The inequality is equivalent to $$\left(\frac{2a}{a+b}\right)^n+\left(\frac{2b}{a+b}\right)^n\ge 2$$ Then, it will be sufficient to prove that $f:(0,2)\to \mathbb{R}$ defined by $f(x)=x^n+(2-x)^n$ has a minimum value of $2$. Taking the derivative of $f$ we have $$f'(x)=nx^{n-1}-n(2-x)^{n-1}$$ So, by ...


2

This looks formal enough, yes. Just make sure to mention that the line $$a+b\leq a+|b|$$ and the line $$b\leq |b|$$ are equivalent (not only that one follows from the other). You can then even use the fact that $b<0$ and $0<|b|$ to prove that last line.


2

If we can show that $(m^2)\times (m^2-1) \times( m^2-2) \times ....\times (m^2-n+2) > m^m $, this will prove the first inequality. \begin{array}\\ P(m, n) &=(m^2)\times (m^2-1) \times( m^2-2) \times ....\times (m^2-n+2)\\ &=\prod_{k=0}^{n-2} (m^2-k)\\ &=\prod_{k=0}^{n-2} (m+\sqrt{k})(m-\sqrt{k})\\ &>\prod_{k=0}^{n-2} (m+\sqrt{k}) ...



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