This is the inequality $$\left(\frac {x^2 + 3}{x}\right) \le 4 $$ This is how I solve it The $x$ in the left side is canceled and $4x$ is subtracted from both sides. $$\not{x} \left (\frac {x^2+3} {\not{x}}\right) \le 4x $$ $$ x^2+3 – 4x \le 4x – 4x $$ $$x^2 -4x + 3 \le 0 […]

Question: let $x_{1},x_{2},\cdots,x_{n}\in \mathbb{R}$,and Assume that the following two sets are equivalent; $$\{[x_{1}],[x_{2}],[x_{3}],\cdots,[x_{n}],\}=\{1,2,3,\cdots,n\},n\ge 2 $$ Find the maximum and minimum of $$\sum_{i=1}^{n-1}[x_{i+1}-x_{i}]$$ where $[x]$ is is the biggest integer not greater than $x$. My idea: since I consider simple case when $n=2$, then $$\{[x_{1}],[x_{2}]\}=\{1,2\}$$ then it is clear $$[x_{1}-x_{2}]_{min}=-2$$ $$[x_{1}-x_{2}]_{max}=1$$ and for general I can’t […]

I know that the sequence $$\displaystyle (1+kx)^\frac{1}{k},$$ where the sequence $\{k_i\}$ converges to zero, converges to $e^x$. I also know the sequence is increasing. How does one show this is increasing? I am interested in neat ways of establishing the inequality, $(1+ax)^\frac{1}{a} \ge (1+bx)^\frac{1}{b}$ if $b \ge a$ rather than the sequence itself.

How to deal with this ? $$\log_{|1 – x|} (x+5)>2 $$ the $|1-x|$ is the base of the logarithm. I tried this below approach but it seems not the complete solution. \begin{align} \frac{\log(x+5)}{\log|1-x|} & > 2\\ {\log(x+5)} & > 2{\log|1-x|}\\ {\log(x+5)} & > {\log|1-x|^2}\\ (x+5) & >|1-x|^2\\ (x+5) & >(1-x)^2\\ (x+5) & >1-2x+x^2\\ x^2-3x-4& < […]

How to prove this inequality ? $$\frac 1{2+a}+\frac 1{2+b}+\frac 1{2+c}\le 1$$ for $a,b,c>0 $ and $a+b+c=\frac 1a+\frac 1b+\frac 1c$. I do not know where to start. I need some idea and advice on this problem.Thanks

The equation is $\sin x < 2x^3$ The steps I’ve taken so far are: $\sin x < 2x^3 $ $\sin x – 2x^3 < 0 $ To solve this I should find when the slope is $ 0 $ so I can find the max and min points and determine the direction of the equation. […]

Hello I have a pretty elementary question but I am a bit confused. I am trying to prove that $$\sum_{k=1}^\infty \frac1{k^2+3k+1} \ge \frac12$$ thanks, Thrasyvoulos

I’m studying Tomas Cormen Algorithms book and solve tasks listed after each chapter. I’m curious about task 1-1. that is right after Chapter #1. The question is: what is the best way to solve: $n\lg(n) \le 10^6$, $n \in \mathbb Z$, $\lg(n) = \log_2(n)$; ? The simplest but longest one is substitution. Are there some […]

$\forall\ x,y,z\in \mathbb{R}$ Show that: $$|x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z|$$ i tired, i notice that $x,y,z$ plays a symmetrical role in the inequality notice also that \begin{align*} |x+y|+|y+z|+|x+z|\leq |x+y+z|+|x|+|y|+|z| & \Longleftrightarrow \\ (|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x+y+z| \end{align*} note that $\forall a,b\in \mathbb{R}\quad |a|-|b|\leq |a+b| $ then $$(|x+y|-|x|)+(|y+z|-|y|)+(|x+z|-|z|) \leq |x|+|y|+|z|$$ i’m stuck here any help would be appreciated!

Let $C$ be a circle of radius $r$ with $n$ points. Prove that there is a point on the circle such that the product of the distances from this point to the other $n$ points is greater than $r^n$. So we seem to be looking at chords that are of length $\leq 2r$. To show […]

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