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

Trouble understanding inequality proved using AM-GM inequality

I am studying this proof from Secrets in Inequalities Vol 1 using the AM-GM inequality to prove this question from the 1998 IMO Shortlist. However, I’m lost on the very first line of the solution. Let x,y,z be positive real numbers such that xyz=1. Prove that x3(1+y)(1+z)+y3(1+z)(1+x)+z3(1+x)(1+y)≥34 Using AM-GM in the following form: x3(1+y)(1+z)+1+y8+1+z8≥3×4. Now, … 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