## Is a generalization of Padoa inequality correct?

Let $$a$$, $$b$$, $$c$$ be sidelengths of a given triangle $$\triangle ABC$$ then

$$(b+c-a)(c+a-b)(a+b-c) \le abc .$$

My question: Is the following generalization of Padoa’s inequality corect?

Let $$a_i>0$$ for $$1\le i\le n$$ and let $$S:=a_1+a_2+….+a_n.$$ Suppose that $$b_i:=S-(n-1)a_i \ge 0\quad\text{ for} \quad 1\le i\le n.$$ Then

$$\prod_{i=1}^n b_i \leq \prod_{1}^{n} a_i .$$

## The optimal asymptotic behavior of the coefficient in the Hardy-Littlewood maximal inequality

It is well-known that for $$f \in L^1(\mathbb{R^n})$$,$$\mu(x \in \mathbb{R^n} | Mf(x) > \lambda) \le \frac{C_n}{\lambda} \int_{\mathbb{R^n}} |f| \mathrm{d\mu}$$, where $$C_n$$ is a constant only depends on $$n$$.

It is easy to see $$C_n \le 2^n$$, but how to determine its optimal asymptotic behavior? For example, does $$C_n$$ bounded in $$n$$? Is $$C_n$$ bounded by polynomial in $$n$$?

## essay on inequality

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essay on inequality

## How to proof the inequality related to Schwarz lemma?

$$f: D\to D$$ $$(D = \{|z| < 1\})$$, is a holomorphic function. How to proof $$\left|\frac{f(z_1)-f(z_2)}{z_1-z_2}\right| < \frac1{1-r^2}, \quad\forall |z_1|,|z_2| \leq r$$ I try using the Schwarz-Pick theorem which gives, $$\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \leq \left|\frac{z_1-z_2}{1-\bar z_1 z_2}\right|$$ then it gives, $$\left|\frac{f(z_1)-f(z_2)}{z_1-z_2}\right| < \frac2{1-r^2}$$ Thanks a lot.

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## Olympiad inequality $\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Big)\sqrt{\frac{\prod_{cyc}(49x-7y+z)}{43^3}}\leq \sqrt{3(x+y+z)}$

I’m interested by the following problem :

Let $$x,y,z>0$$ with $$49x-7y+z>0$$, $$49y-7z+x>0$$ , $$49z-7x+y>0$$ then we have : $$\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Big)\sqrt{\frac{\prod_{cyc}(49x-7y+z)}{43^3}}\leq \sqrt{3(x+y+z)}$$

I have tested this inequality with Pari-Gp and it seems to be okay . Furthermore I think we can use the $$uvw$$‘s method (because the equality case comes when $$x=y=z$$) but I don’t see how now . I have a ugly proof using derivative (the inequality can be reduce to a two variable inequality) but it’s too long to be explain here . The inequality is too precise to use Jensen’s or Slater’s inequality here. Finally I have also tested brut force but I can’t find an interesting irreductible factorization .

If you have a hint it would be nice .

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## Markov inequality: More precise bound?

Given a random variable X, we know that $$P[X\geq A] = 1$$. By Markov inequality, we obtain that $$E[X]\geq A$$. Or, in other words, $$E[X] = A + \lambda,\,\,\lambda\geq 0$$. Is there any way I can more precisely characterize the $$\lambda$$? E.g., if I know the variance of $$X$$? Or applying some other bounds, less conservative than Markov’s.

## Kraft’s inequality for Huffman coding

Is it true to say that Huffman codes always satisfy Kraft’s inequality with strict equality?

At first, I thought the statement is wrong since for Huffman coding there is no assumptions on the probabilities. However, when I tried evaluation the Kraft inequality for several Huffman codes, I realized that it’s got nothing to do with the probabilities. It’s the algorithm.

I looked for a nice mathematical proof for this, but couldn’t find any. Would appreciate your help.

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## How prove this inequality maybe is Karamata inequality

The Following problem is from the Geometry problem,let $$x_{i}>0(i=1,2,\cdots,n)$$,and such $$x_{1}+x_{2}+\cdots+x_{n}=\pi$$ show that $$\dfrac{\sin{x_{1}}\sin{x_{2}}\cdots\sin{x_{n}}}{\sin{(x_{1}+x_{2})}\sin{(x_{2}+x_{3})}\cdots\sin{(x_{n}+x_{1})}}\le\left(\dfrac{\sin{\frac{\pi}{n}}}{\sin{\frac{2\pi}{n}}}\right)^n$$I tried everything, but I failed. consider $$f(x)=\ln{\sin{x}},0,since $$f”(x)=-csc^2{x}<0$$ it suffices to prove that $$f(x_{1}+x_{2})+f(x_{2}+x_{3})+\cdots+f(x_{n}+x_{1})+nf(\dfrac{\pi}{n})\ge f(x_{1})+f(x_{2})+\cdots+f(x_{n})+nf(\dfrac{2\pi}{n})$$ or $$f(x_{1}+x_{2})+f(x_{2}+x_{3})+\cdots+f(x_{n}+x_{1})+nf(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n})\ge f(x_{1})+f(x_{2})+\cdots+f(x_{n})+nf(\dfrac{2(x_{1}+x_{2}+\cdots+x_{n})}{n})$$ in other words,if $$f”(x)\le 0$$,we can prove following inequality? $$f(x_{1}+x_{2})+f(x_{2}+x_{3})+\cdots+f(x_{n}+x_{1})+nf(\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n})\ge f(x_{1})+f(x_{2})+\cdots+f(x_{n})+nf(\dfrac{2(x_{1}+x_{2}+\cdots+x_{n})}{n})?$$ has anyone studied the work?Thanks!

## Distance function to the boundary and Harnack inequality

Suppose $$\Omega \subset \mathbb{R}^d$$ be a domain, and let $$\rho(x) = \mathrm{dist} (x, \partial \Omega)$$ be the distance function to the boundary of $$\Omega$$. I want to know for which domains $$\rho$$ satisfies a Harnack type inequality. Harnack inequality says that $$\sup _{x \in B} \rho (x) \leq C \inf _{x \in B} \rho(x)$$ on a ball $$B= B(a,r), a \in \Omega$$, and $$C$$ is a constant depend on $$B$$. It is known that harmonic functions satisfy Harnack inequality. Is it enough if $$\Omega$$ satisfy regularity property (e.g.,if it is a Lipschitz or a NTA-domain)? What about boundary Harnack inequality?

## About one integral inequality under a constraint

Set $$\phi (x) = u(x)+iv(x)$$, $$x=(x_1,…,x_N)$$, a $$T$$-periodic function in $$H^1_{loc}(\mathbb{R}^N)$$, that is $$\phi (x) = \phi (x_1 + T ,…, X_N + T)$$ for all $$x$$ and where $$u = Re\ \phi$$ and $$v = Im \ \phi$$. I wonder if $$\int_{[0,T]^N}\frac{u_{x_1}^2}{2} + \frac{v_{x_1}}{\sqrt{2}}(u-1)^2\ dx > 0$$ for functions satisfying $$\int_{[0,T]^N} v_{x_1}(u-1)\ dx = \varepsilon > 0$$ for some small and fixed $$\varepsilon > 0$$. Any idea or help to show this would be appreciate! I am stucked. Thank you in advance.