Tagged Questions

For questions about or involving the absolute value function.

1answer
22 views

Proving an inequality given some conditions.

I would like to prove the statement: If $|a| > |b|$, with $a > 0$, where $a$ and $b$ are real numbers, then $|a + a^{2}| > |b + b^{2}|$. I am fairly certain that this claim is true. ...
0answers
38 views

Cases in which a certain inequality holds true.

I previously asked this question on this forum, and have been demonstrated counterexamples to the claim that $|a| > |b|$ implies $\big|\frac{b+b^{2}}{a+a^{2}}\big| < 1$, which I had previously ...
3answers
40 views

$f(x)=x+\frac{1}{e^x+1}$. Prove that for any $x,y$ : $|f(x)-f(y)|\leq|x-y|$

I feel like this question is related to the Mean value theorem, but the absolute value interferes with it. I get to: $$\frac{|f(x)-f(y)|}{|x-y|}\leq 1$$ And from there I want to prove that the ...
3answers
73 views

Proving or disproving that an inequality implies another inequality.

I am wondering if $|a| > |b|$ implies $|\frac{b+b^{2}}{a+a^{2}}| < 1$, where $a$ and $b$ are real numbers. I have tested numerically with many cases and I have found this to be true in all of my ...
1answer
56 views

Definition of absolute value of complex number

The definition I see everywhere for the absolute value of a complex number is: Let $z=x+iy$ then $|z| := \sqrt{x^2+y^2}$. But the square root operation is multivalued in complex analysis. So while ...
3answers
65 views

Prove that a specific inequality holds

Let $n \in \mathbb{N}$. Let $z_1, \ldots, z_n$ and $w_1, \ldots, w_n$ be complex numbers such that $$\sum_{j = 1}^n |w_j|^2 \leq 1$$ and $$\left| \sum_{j = 1}^n z_j w_j \right| \leq 1$$ Show that ...
2answers
63 views

Can you explain why $x>\frac 23$ is not a solution to this inequality

$|2x + 2| + |x - 1| > 3$ why can't $x>\frac 23$ be a part of the solution? Thanks for your help!
0answers
20 views

Univalent triangle inequality [duplicate]

$|Z_1| = | \frac{v(1+\alpha) + \sqrt{v^2(1+\alpha)^2-4\alpha}}{2}|$ Triangle inequality |x+y|=|x|+|y| Where x= $\frac{v(1+\alpha)}{2}$ and $y= \frac{\sqrt{v^2(1+\alpha)^2-4\alpha}}{2}$ I've been ...
1answer
40 views