# Tag Info

### Let $x,y,z\in[0,1]$. Find the maximum value of $\sqrt{|x-y|}+\sqrt{|y-z|}+\sqrt{|z-x|}$.

WLOG fix $x\geq y\geq z$. Then our expression $E$ is $$E=\sqrt{x-y}+\sqrt{y-z}+\sqrt{x-z}$$ Now, note that as $x$ increases, so does $E$ and so for the maximum value of $E$ we should have $x=1$. ...
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### An interesting recurrent equality, possibly easier to solve in its differential form?

Your question comes in two variants. One is discrete. The other continuous. The discrete variant asks about the equation $$f(n+1) - f(n) = c f(n)\sum_{m=0}^n f(m). \tag{1}$$ A nicer version of this ...
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### Is there a shorter or more trivial way to prove that $x > \cos (x)-\cos (2 x)$ holds for all $x>0$?

This can be seen geometrically. Let $P = (\cos x, \sin x)$ and $Q = (\cos 2x, \sin 2x)$; then arc $PQ$ on the unit circle has length $x$. Its projection onto the horizontal axis is an interval ...
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### Is there a shorter or more trivial way to prove that $x > \cos (x)-\cos (2 x)$ holds for all $x>0$?

If $x>2$, the statement is trivial. For $0<x\leq 2,$ one has $$x>2\sin\left(\frac x2\right)\geq 2\sin\left(\frac{3x}2\right)\sin\left(\frac x2\right)=\cos(x)-\cos(2x).$$
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### Inequality involving integral of 1/log(t)

By substituting $u = \frac{\log t}{\log x}$, we find that $$\int_{x}^{x^2} \frac{1}{\log t} \, \mathrm{d}t =\int_{1}^{2} \frac{x^u}{u} \, \mathrm{d}u.$$ Now use the fact that $x \leq x^u \leq x^2$ ...
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### Is there a shorter or more trivial way to prove that $x > \cos (x)-\cos (2 x)$ holds for all $x>0$?

For $x > 0$ is $$\cos(x)-\cos(2x) = \int_x^{2x} \sin(t) \, dt < \int_x^{2x} 1 \, dt = x \, .$$ Strict inequality between the integrals holds because $\sin(t) < 1$ for “almost all” $t$.
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### An unusual inequality problem involving real exponential power

The form $4\omega^\alpha+\omega^{n-4\alpha}$ strongly suggests it's AGM time: $$\frac{4\omega^\alpha+\omega^{n-4\alpha}}{5}\ge\sqrt[5]{\omega^{4\alpha}\omega^{n-4\alpha}} = \omega^{n/5}.$$ So we ...
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### Prove $\sqrt{x-1} + \sqrt{y-1} \le xy, x \ge 1, y \ge 1$

you can use schwartz inequality directly we know ;$a_1 b_1+a_2b_2 \leq \sqrt{a_{1}^2+a_{2}^2} \sqrt{b_{1}^2+b_{2}^2}$ just put $a_1 =\sqrt{x-1}$ $a_2 =1$ $b_1 =1$ and $b_2 =\sqrt{y-1}$ so you will get ...
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### If $−1<x<4$ then determine $a$ and $b$ in $a<2x+3<b$.

Yes, you did do something wrong: abuse notation without understanding what is based on, and misinterpreting the result that should be obtained. $a$ is in the interval $(-\infty,1]$ but that doesn't ...
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### An interesting recurrent equality, possibly easier to solve in its differential form?

Being a combination of product and sum, it looks difficult that there might be a closed form. The best I can suggest is to make the substitution $$f(n) = 2^{g(n)}$$ I am using $2$ as a base because ...
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Since the inequality in question is used primarily to derive bounds in $L_p$, namely Clarkson's inequality for $1<p<2$: $$\Big\|\frac{f+g}{2}\Big\|^{p'}_p+\Big\|\frac{f-g}{2}\Big\|^{p'}_p\leq \... • 26.4k 1 vote Accepted ### Let X be an exponential random variable and \Bbb{P}(X \in A) < \Bbb{P}(X \in B). Is \Bbb{P}(aX \in A) < \Bbb{P}(aX \in B) for a > 0? Not necessarily. Take for instance A=(x,+\infty) and B=(0,y) for any x,y>0 such that 1-\exp(-x)>\exp(-y), so that \mathbb P(X\in A)<\mathbb P(Y\in B). For instance, x=\ln(3) and ... • 3,256 1 vote ### Bounding spectral radius of special matrix (extension of the extension) This is not an answer, but some thoughts. Consider the case when D_{2} has only one positive diagonal entry, and all other diagonal entries equal to 0 -- similar to what we did here. First, as in ... • 326 1 vote ### Solution verification: is that quantity bounded from below? The claim only holds for N \geq 0 and p \geq 2. To see why you need N \geq 0, note that if N < 0 then$$\frac{|t|^p}{s+1}e^{|t|^{p'}} - 2NF(t) = \frac{|t|^p}{s+1}e^{|t|^{p'}} + CF(t) for \$...

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