## System of Equations: any solutions at all?

I am looking for any complex number solutions to the system of equations: \begin{align} |a|^2+|b|^2+|c|^2&=\frac13 \\ \bar{a}b+a\bar{c}+\bar{b}c&=\frac16 (2+\sqrt{3}i). \end{align} Note I put inequality in the tags as I imagine it is an inequality that shows that this has no solutions (as I suspect is the case). This is connected to my other question… I have … Read more

## Verification for this proof

Sorry guys about the verification questions but it’s near the end of the semester and I am very sheepish about making mistakes especially because real analysis is a very important course it’s only the underlying theory of calculus. This problem may seem rudimentary but we were given a sample test by our professor and it … Read more

## Concluding three statements regarding $3$ real numbers.

$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$ Conclusion $I.)\ 1<b<3$ Conclusion $II.)\ 2<a<3$ Conclusion $III.)\ 0<c<1$ Options By the given statements $\color{green}{a.)\ \text{Only conclusion$I$can be derived}}$. $b.)\$ Only conclusion $II$ can be derived. $c.)\$ Only conclusion $III$ can be derived. $d.)\$ Conclusions $I,\ II,\ III$ can be derived. $e.)\$ None … Read more

## Prove this inequality (a1a2⋯an)√1−an+1+√n−1⋅an+1<√n(a_{1}a_{2}\cdots a_{n})\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}<\sqrt{n}

Assmue that ai∈(0,1),i=1,2,3,⋯,n,show that (a1a2⋯an)√1−an+1+√n−1⋅an+1<√n, I’ve tried many things but all have failed. Answer If u∈(0,1) then, by Cauchy-Schwarz, √1nu+√1−1n(1−u2)≤√u2+(1−u2)2<√u2+(1−u2)=1 Take u=√1−an+1 and rearrange to get √1−an+1+√n−1an+1<√n Introducing a1⋯an on the first term on the left only makes the left smaller. AttributionSource : Link , Question Author : Community , Answer Author : Community

## Prove that ∑cyc√a+bc≥2∑cyc√ca+b\sum\limits_{cyc}\sqrt{\frac{a+b}{c}}\ge2\sum\limits_{cyc}\sqrt{\frac{c}{a+b}}

Let a,b,c be positive numbers. Then we need to prove √a+bc+√b+ca+√c+ab≥2(√ca+b+√ab+c+√bc+a). I have an idea to set x=ab+c, y=bc+a,z=ca+b then 11+x+11+y+11+z=2 and we need to prove 1√x+1√y+1√z≥2(√x+√y+√z) But I could not go further. Answer It is a consequence of Chebychev’s inequality: ∑cyc√a+bc≥2∑cyc√ca+b⟺∑cyca+b−2c√c(a+b)≥0 Since the a+b−2c and 1√c(a+b) are ordered in the same way, we can … Read more

## Finding a function satisfying a certain inequality

This is a continuation of this post where I tried to find a function f(n) that would satisfy the induction step of an inductive argument and it was shown that such function does not exist. Trying to fix the problem I’ve come up with a stronger inductive argument that now requires finding a more elaborate … Read more

## Proving that $\sum_{i=1}^n\frac{1}{i^2}<2-\frac1n$ for $n>1$ by induction [duplicate]

This question already has answers here: Proving $1+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{n^2}\leq 2-\frac{1}{n}$ for all $n\geq 2$ by induction (5 answers) Closed 3 years ago. Prove by induction that $1 + \frac {1}{4} + \frac {1}{9} + … +\frac {1}{n^2} < 2 – \frac{1}{n}$ for all $n>1$ I got up to using the inductive hypothesis to prove that … Read more

## Modifying a Textbook Theorem in Real Analysis

In my real analysis book, there is a theorem which states that if we let x,y∈R such that x≤y+ϵ for all ϵ>0, then x≤y. My question is, if we instead have k⋅ϵ where k is a nonzero constant, will the inequality still be true? That is, if x≤y+k⋅ϵ, then x≤y? Or must we have k⋅ϵ>0? … Read more

## Why two Inequalities are true

A is a subset of real numbers. Consider the set A of all real numbers x such that x2≤2. This set is nonempty and bounded from above, for example by 2. Call the sup. Then y^2 = 2, because the other possibilities y^2 \lt 2 and y^2 \gt 2 both lead to a contradiction. Assume … Read more

## Basic algebra inequality mistake (I probably didn’t break algebra)

I’m not sure what I have done here. I’m guessing it’s something to do with the absolute values but I really have no idea what it is. y>1yy>1yy2>11>1y2|y|>11>1|y|1|y|>1↓↘↙1<|1y|<1 Answer The error is on the left. You went from |y|>1 to |1y|>1 This is incorrect. You should have divided both sides by |y|, leaving 1>1|y| This … Read more