Articles of inequality

On the arctangent inequality.

As in the title, how can I prove $$ \frac{\arctan(x)}{x}\geq\frac{1}{2} $$ for $x\in(0,1]$? I think I can say: $$\frac{\arctan(x)}{x}$$ is monotonically decreasing in the interval, so its value is greater than the value in $1$, which is $\frac{\pi}{4}$, greater than $\frac{1}{2}$. There exists some elegant proof of this simple inequality, maybe using series expansions?

Subtracting positive numbers from denominator in an inequality.

When we have the following inequality: $$\frac{a}{b+c} \ge \frac{d}{e+c},$$ with $a,b,c,d,e \in \mathbb R_{\ge 0}$ Then it seems to hold that $$\frac{a}{b} \ge \frac{d}{e},$$ Is this correct? Does it also work in the other direction (iff)?

Remarks on a Previous Post: Elementary proof of $n>\frac{p_n}{\ln p_n}$

Recently I have been reading this post and I have noted something significant in the fake argument. As one can easily see that the basic idea behind the argument had been to show that the sequence $x_n=\dfrac{\pi(n)\ln n}{n}$ is strictly decreasing for all $n$. But, notice that the Prime Number Theorem is equivalent to the […]

Prove that $ \sqrt{\sum_{j=1}^{n}a_j^2} \le \sum_{j=1}^{n}|a_j|$, for all $a_i \in \mathbb{R}, i=\{1,2,…,n\}$

As the title says, prove that $ \sqrt{\sum_{j=1}^{n}a_j^2} \le \sum_{j=1}^{n}|a_j|$, for all $a_i \in \mathbb{R}, i=\{1,2,…,n\}$. I can solve for the case of n=2, but kind of stuck while proving the n-terms version. The attempt is as follow: $\left(\sqrt{\sum_{j=1}^{n}a_j^2}\right)^2 = \sum_{j=1}^n {a_j}^2 \le |\sum_{j=1}^n {a_j}^2| \le \sum_{j=1}^n |a_j|^2 \mbox{by triangle inequality.}$ Alright, I am stuck […]

If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, and $k=1$, does it follow that $\frac{\sigma(n^2)}{n^2} \geq 2 – \frac{5}{3q}$?

Let $\sigma=\sigma_{1}$ be the classical sum-of-divisors function. If $N = q^k n^2$ is an odd perfect number with Euler prime $q$ (i.e., $q$ satisfies $q \equiv k \equiv 1 \pmod 4$), and $k=1$, does it follow that $$\frac{\sigma(n^2)}{n^2} \geq 2 – \frac{5}{3q}?$$ Note that, since the Euler prime $q$ satisfies $q \equiv 1 \pmod 4$, […]

Is $x^t$ subadditive for $t \in $?

I feel like I have seen this many times, that $(x+y)^t \leq x^t + y^t$ when $t \in [0,1]$, but I don’t remember ever proving it. I feel like it surely must be true. Can someone lead me in the right direction on this one?

An inequality of integrals

Let $f\in C^1([a,b])$ with $f(a)=0$. How can I show that there exists a positive constant $M$ independent of $f$ such that $\int^b_a|f(x)|^2dx\leq M\int^b_a|f^\prime(x)|^2dx$?

The ratio $\frac{u(z_2)}{u(z_1)}$ for positive harmonic functions is uniformly bounded on compact sets

I want to prove the following: If $E$ is a compact set in a region $\Omega \subset \mathbb C$, prove that there exists a constant $M$, depending only on $E$ and $\Omega$, such that every positive harmonic function $u(z)$ in $\Omega$ satisfies $u(z_2) \leq M u(z_1)$ for any two points $z_1, z_2 \in E$. This […]

Mean Power Inequality

Most of the proofs of mean power inequality are based on jensen’s inequality. Can the mean power inequality be prooved without use of one? Mean Power Inequality: http://www.artofproblemsolving.com/Wiki/index.php/Power_Mean_Inequality

How prove this $|x_{p}-y_{q}|>0$

let $$x_{1}=\dfrac{1}{8},x_{n+1}=x_{n}+x^2_{n},y_{1}=\dfrac{1}{10},y_{n+1}=y_{n}+y^2_{n}$$ show that: for any $p,q\in N^{+}$ we have $$|x_{p}-y_{q}|>0$$