Is a generalization of Padoa inequality correct?

Padoa’s inequality is named after Alessandro Padoa (1868-1937):

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.

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 .

Thanks in advance .

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.

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<x<\pi$ ,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.