Articles of inequality

Prove that $y-x < \delta$

In Hardy’s Pure Mathematics it says if $x^2<2$, $ \ \ y^2>2$, $ \ \ 2-x^2 < \delta$ and $y^2 – 2 < \delta$, then $y-x<\delta$. I added the last two inequalities to get $(y+x)(y-x)<2\delta$. How do I proceed from here?

$\sum a+b+c \sum \frac{ab+bc+ca}{a+b+c} \sum \frac{abc}{ab+bc+ca} \leq \sum a \sum b \sum c$

Let $a_i,b_i,c_i$ be $>0$ ($1 \leq i \leq n$). Then we have $$ \sum (a_i+b_i+c_i) \sum \frac{a_i b_i + b_i c_i + c_i a_i}{a_i+b_i+c_i} \sum \frac{a_i b_i c_i}{a_i b_i+b_i c_i + c_i a_i} \leq \sum a_i \sum b_i \sum c_i $$ This is problem #68 in Hardy, Polya and Littlewood’s Inequality. It can be proved […]

Prove the inequality for composite numbers

Is it true that $c_m+c_n$ $>$ $c_{m+n}$ for all $m$, $n$ $\in$ $\mathbb{N}$? Though the result seems true, I can’t get a solution. Even the bounds on $c_n$ obtained from Prime Number Theorem isn’t helping me. Is there any way to prove it?

Etemadi's inequality

In another post an inequality referred to as “Etemadi’s Inequality” is mentioned twice – in the original post as well as in the answer. However, the contexts of usage are such as to raise the question whether the inequality intended by the users (ziT and saz, respectively) is the inequality that goes by the same […]

Equality in the Schwarz-Pick theorem implies function is a linear fractional?

Part of the Schwarz-Pick Theorem states that for an analytic automorphism of the unit disk, then $$ \frac{|f'(z)|}{1+|f(z)|^2}\leq\frac{1}{1-|z|^2}. $$ In the wikipedia article of the Schwarz-Pick theorem, it is mentioned that if equality holds, then $f$ is a Moebius transformation on the unit disk without proof. Is there a proof of this detail? Thank you.

Trigonometric Triangle Equality

$A, B, C$ are the angles of a triangle then $tan^2(A/2)+tan^2(B/2)+tan^2(C/2)$ is always greater than what integral value.

Prove $x \geq \sin x$ on $$

As the title says.. it says to use the mean value theorem but I don’t see how that’s applicable. Thank you

Prove this inequality $\frac{1}{xy+z}+\frac{1}{yz+x}+\frac{1}{zx+y}\le\frac{1}{2}$

Let $x,y,z>0$ and such $xy+yz+xz\ge 2(x+y+z)$,show that $$\dfrac{1}{xy+z}+\dfrac{1}{yz+x}+\dfrac{1}{zx+y}\le\dfrac{1}{2}$$

absolute value inequalities

When answer this kind of inequality $|2x^2-5x+2| < |x+1|$ I am testing the four combinations when both side are +, one is + and the other is – and the opposite and when they are both -. When I check the negative options, I need to flip the inequality sign? Thanks

Friedrichs's inequality?

Friedrichs’s second inequality is stated as follows(see For all $\mathbf{u} \in H^1(\Omega)^2$ satisfying either $\mathbf{n}\cdot\mathbf{u} = 0$ or $\mathbf{n} \times \mathbf{u} = \mathbf{0}$ on $\partial\Omega$ where $\Omega$ is a simply connected domain, then $$ \|\mathbf{u}\|_1 \le C_1 (\|\nabla\cdot\mathbf{u}\|_0 + \|\nabla\times\mathbf{u}\|_0). $$ My question is that if the boundary condition is satisfied only on the […]