## Solving an Inequality on Integers II

Here is an exercise similar to the old one:

How do I show 2.?

## Python – Monte Carlo method for checking the triangle inequality

This is a code to find inequality $$abc> 2 ( text {area of ​​the triangle})$$
Where $$a$$, $$b$$ and $$c$$ are the sides of a triangle.

The following and the corresponding code work as intended. However, they take a long time. Performance can be improved without reducing max_tries?

import random
import math
hn=100000000
num=0
max_tries=10000000
tries = 0
while tries != max_tries:
a= random.randint(1,hn)
b= random.randint(1,hn)
c= random.randint(1,hn)
if a+b>c and c+b>a and a+c>b :
s=(a+b+c)/2
tries= tries +1
k =math.sqrt(s*(s-a)*(s-b)*(s-c))
if a*b*c> 2*k:
num= num+1

print(num/max_tries)


## Geometry – help with scalar product inequality

To let $$(a_1, a_2), (w, z), (u, v) in mathbb {R} ^ 2$$ so that $$vert (w, z) vert leq 1$$ and $$vert (u, v) vert geq R$$, Where $$R$$ is a constant greater than $$1$$.

It's true that
$$begin {equation} vert left langle (a_1, a_2), (w, z) right rangle vert leq vert left langle (a_1, a_2), (u, v) right rangle vert? end {equation}$$
$$begin {equation} vert left langle (a_1, a_2), (w, z) right langle vert leq vert (a_1, a_2) vert leq vert (a_1, a_2) cdot R vert leq vert left langle (a_1, a_2), (u, v) right rangle vert end {equation}$$

## Calculus – The inequality formula to prove $x * 2$ is continuous over a fixed finite interval [a, b]

This is discussed in Courant's introduction to Calculus and Analysis I. $$f (x) = x ^ 2$$ is continuous over a fixed finite interval (a, b).

It is written as:
$$| f (x) – f (x_ {0}) | = | x ^ 2 – x_ {0} ^ 2 | = | x – x_ {0} || x + x_ {0} | leqslant 2 | x – x_ {0} | (| b | + | a |) <2 sigma (| b | + | a |) <2 epsilon$$

I have no idea what inequality sentences the author used:

1. $$| x – x_ {0} || x + x_ {0} | leqslant 2 | x – x_ {0} | (| b | + | a |)$$

2. $$2 | x – x_ {0} | (| b | + | a |) <2 sigma (| b | + | a |)$$

## Probability distributions – How can I use the Chernoff inequality to find an upper bound?

Given a number of random variables, so for each $$i$$ $$in$$ $$mathbb N$$::
$$Pr (X_i = -9) = 1/10$$
$$Pr (X_i = 1) = 9/10$$

And I define for everyone $$n$$ $$in$$ $$mathbb N$$:: $$S_n = sum ^ n_ {i = 1} X_i$$.

How can I use chernoff inequalitie to find an upper limit for? $$Pr (| S_n |> 10 sqrt {n})$$ ?

## Inequality – Proof that x ^ 4 + y ^ 4 + z ^ 4> = x ^ 3y + y ^ 3z + z ^ 3x with weighted Am-Gm

But avoid

• Make statements based on opinions; Support them with references or personal experiences.

Use MathJax to format equations. MathJax reference.

## nt.number theory – inequality in Iwaniec-Kowalski

I read about Dirichlet polynomials in the book Analytic Number Theory by the authors mentioned. Can anyone justify the following inequality? Assume that $$a (n), b (m)$$ are sequences of non-negative numbers, so $$a (n) = 0$$ big enough for everyone $$n$$ and $$b (m) = 0$$ big enough for everyone $$m$$. How do you prove that? $$sum _ { substack {n, m, n_1, m_1 in mathbb N \ nm = n_1 m_1}} a (n) b (m) a (n_1) b (m_1) leq sum_n sum_m a (n) ^ 2 b (m) ^ 2 tau (nm)?$$ Here $$dau$$ is the function that counts the number of positive integer divisors. This is according to equation (9.7) on page 231.

## Complex inequality: find the maximum value.

Given that
| a | = 1
| b-2 | = 3
| c-5 | = 6
Find the maximum value of | 2a-3b-4c |
where a, b, c are complex numbers.

## Probability – An inequality for the mgf using the Jensen inequality

Given non-negative random variables $$X_1, X_2, …$$ how to show that
$$mathbb {E} exp (t max borders_ {1 leq i leq n} X_i) leq sum borders_ {1 leq i leq n} mathbb {E} exp (tX_i ).$$
$$max limit_ {1 leq i leq n} X_i leq sum limit_ {1 leq i leq n} X_i$$
To let $$T in B (E)$$ be a limited linear operator on the Banach grid $$E$$. Consider his addition $$T: E ^ * rightarrow E ^ *$$ defined in the usual way as $$T ^ * (y (x)): = y (Tx)$$ With $$y$$ a limited function in $$E ^ *$$.
Now leave $$mu in E ^ *$$ be an eigenvector of $$T$$ whose intrinsic value is $$alpha neq 0$$and define $$nu: = T ^ * R (r_1, T ^ *) | mu |$$ With $$r_1> rho (T)$$ ($$R$$ denotes the resolving operator and $$rho$$ the spectral radius of $$T$$).
How to show that $$T ^ * nu le r_1 nu$$?