Articles of inequality

Proving: $\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.$

$$\begin{equation}\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0\end{equation}$$ I found this inequality in this paper: http://ajmaa.org/RGMIA/papers/v7n2/pade.pdf (Equation (3)). How exactly can I prove it? I tried induction but to no avail…

Show that $\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$.

Show that $$\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$. I tried many different ways to expand left side and estimate it but always got stuck at some point.

How prove this $|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\frac{1}{20n^3}$

Prove that $$|\{n\sqrt{3}\}-\{n\sqrt{2}\}|>\dfrac{1}{20n^3}$$ let $t=\{n\sqrt{2}\}-\{n\sqrt{3}\}$ and $k=[n\sqrt{3}]-[n\sqrt{2}]$ then we have $$t=k-(\sqrt{3}-\sqrt{2})n=k-\sqrt{5-2\sqrt{6}}n\neq 0$$ so \begin{align*} t&=\dfrac{(k-(\sqrt{3}-\sqrt{2})n)(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}{(k-(\sqrt{3}+\sqrt{2})n)(k+(\sqrt{3}-\sqrt{2})n)(k+(\sqrt{3}+\sqrt{2})n)}\\ &=\dfrac{k^4-10k^2n^2+n^4}{(t-2\sqrt{2}n)(t+2(\sqrt{3}-\sqrt{2})n)(t+2\sqrt{3}n)} \end{align*} notice that $$|t-2\sqrt{2}n|\le 2\sqrt{2}n+\dfrac{1}{20}\le(2\sqrt{2}+\dfrac{1}{20})n$$ $$|t+2(\sqrt{3}-\sqrt{2})n|\le2(\sqrt{3}-\sqrt{2})n+\dfrac{1}{20}\le(2\sqrt{3}-2\sqrt{2}+\dfrac{1}{20})n$$ $$|t+2\sqrt{3}n|\le2\sqrt{3}n+\dfrac{1}{20}\le(2\sqrt{3}+\dfrac{1}{20})n$$ so $$|t|\ge\dfrac{1}{(2\sqrt{2}+\dfrac{1}{20})(2\sqrt{3}+2\sqrt{2}-\dfrac{1}{20})(2\sqrt{3}+\dfrac{1}{20})n^3}>\dfrac{1}{7n^3}$$ so $$|t|\ge\min{\left(\dfrac{1}{20},\dfrac{1}{7n^3}\right)}\ge\dfrac{1}{20n^3}$$ \ This post https://math.stackexchange.com/questions/465419/how-prove-this-t-2-sqrt2n-le2-sqrt2n-frac120 is not true? so This methods is wrong, so How prove it? Thank you

Proof for Inequality

Can somebody tell me what is the name of the inequality: \begin{equation} \sum_{t=1}^T \frac{1}{\sqrt{t}} \leq 2\sqrt{T} \end{equation} or any hint/link how to prove above? Thanks.

Prove that $-3/2\leq\cos a + \cos b + \cos c\leq 3$?

Given 3 non-null vectors $v,u,w$ and angles $a=(u,v), b=(u,w), c=(v,w)$, Prove that $-3/2\leq\cos a + \cos b + \cos c\leq 3$. I’ve managed to prove that: $\cos a + \cos b + \cos c\leq 3$ basically arguing that $\cos \theta$ is bounded by $-1,1$ using the inequality of Cauchy-Schwarz. I reasoned that as the maximum […]

Inequality with two sequences

Let $x_1,\dots,x_n$ and $y_1,\dots,y_n$ be two increasing sequences of nonnegative real numbers with $x_i\leq y_i$ for all $i$. Is there a constant $c>0$ (independent of $n$) for which there exists some $r\geq 0$ (possibly dependent on $n$ and the sequences) such that $$\sum_{i: x_i\leq r\leq y_i}y_i+\sum_{i:x_i\geq r} x_i\geq c\sum_{i=1}^n y_i?$$ This is a discrete version […]

If $N=q^k n^2$ is an odd perfect number and $q = k$, why does this bound not imply $q > 5$?

Let $\mathbb{N}$ denote the set of natural numbers (i.e., positive integers). A number $N \in \mathbb{N}$ is said to be perfect if $\sigma(N)=2N$, where $\sigma=\sigma_{1}$ is the classical sum of divisors. For example, $\sigma(6)=1+2+3+6=2\cdot{6}$, so that $6$ is perfect. (Note that $6$ is even.) Denote the abundancy index of $x \in \mathbb{N}$ as $I(x)=\sigma(x)/x$. Euler […]

Prove $F(n) < 2^n$

Consider the Fibonacci function $\large{F(n)}$, which is defined such that $F(1) = 1, F(2) = 1$, and $F(n) = F(n−2)+F(n−1)$ for $n > 2$ I know that I should do it using mathematical induction but I don’t know how to approach it. Can anyone help me prove $F(n) < 2^n$. Thank so much

How can I prove that $\left|\sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i}\right| \leq \binom{n}{r}$?

This is a conjecture: How can I prove that \begin{equation} \left|\sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i}\right| \leq \binom{n}{r} \end{equation} for $0\leq a \leq n$, $0\leq r \leq n$ and $n,r,a \in \mathbb{N}$ ?

Inequality. $\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \geq \frac{(x+y+z)^3}{18}$

Let $x,y$ and $z$ be positive numbers such that $xy+yz+zx=1$. Prove that (using Hölder’s inequality) : $$\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \geq \frac{(x+y+z)^3}{18}$$ Thanks 🙂 What I try: $$\left(\frac{x^3}{1+9y^2xz}+\frac{y^3}{1+9z^2yx}+\frac{z^3}{1+9x^2yz} \right)\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right) \geq \left(\sum_{x,y,z}{\left(\sqrt[3]{\frac{x^3}{1+9y^2xz}\cdot\left(1+9y^2xz\right) \cdot 1}\right)}\right)^{3}=\left(\sum_{x,y,z}{x}\right)^{3}.$$ So we have to prove that : $$\large\frac{\left(\sum_{x,y,z}{x}\right)^{3}}{\left(1+9xy^2z+1+9xyz^2+1+9xyz^2\right)\left(1+1+1\right)} \geq \frac{(x+y+z)^3}{18} $$ or $$3\cdot \left(3+9xyz\left( x+y+z\right)\right) \leq 18 \Leftrightarrow$$ $$xyz\left(x+y+z\right) \leq \frac{1}{3},$$ but I don’t know […]