This videoes are pretty awesome to develop the basics of combinatorics.

https://m.youtube.com/playlist?list=PLx ... =mv-google

## Search found 12 matches

- Sat Nov 26, 2016 9:31 pm
- Forum: Combinatorics
- Topic: 'BASIC THINGS OF COMBINATORICS
- Replies:
**2** - Views:
**1398**

- Sat Nov 12, 2016 6:10 pm
- Forum: Number Theory
- Topic: SSC Mock'16 : Problem 1
- Replies:
**0** - Views:
**867**

### SSC Mock'16 : Problem 1

Well, this might be easy but still an interesting one to me.

Find all finite sets of positive integers with at least two elements such that for any two numbers $a,b$ belonging to the set with $a>b$, the number $\dfrac{b^2}{a-b}$ belongs to the set, too.

Find all finite sets of positive integers with at least two elements such that for any two numbers $a,b$ belonging to the set with $a>b$, the number $\dfrac{b^2}{a-b}$ belongs to the set, too.

- Sun Nov 06, 2016 10:37 pm
- Forum: Algebra
- Topic: Iran TST 2012,Exam2,Day2,P4
- Replies:
**1** - Views:
**1969**

### Re: Iran TST 2012,Exam2,Day2,P4

My solution Let $a \ge b \ge c > 0$. Then by applying chebyshev, we get, \[ \sum_{cyc}^{} \dfrac{a\sqrt{a}}{bc} \ge \dfrac{1}{3} \left( \sum_{cyc}^{} \dfrac{a}{bc} \right)\left( \sum_{cyc}^{} \sqrt{a} \right) \] Then it's enough to prove that \begin{align*} \dfrac{1}{3} \left( \sum_{cyc} \dfrac{a}{b...

- Sat Nov 05, 2016 11:48 am
- Forum: Algebra
- Topic: A Geometric Inequality
- Replies:
**1** - Views:
**1858**

### A Geometric Inequality

Let $a,b,c$ be the sides of $\triangle ABC$ and $x$ be any non-negative real number. Prove that,

\[ a^x \cos A+ b^x \cos B + c^x \cos C

\leq \dfrac{1}{2}(a^x+b^x+c^x) . \]

\[ a^x \cos A+ b^x \cos B + c^x \cos C

\leq \dfrac{1}{2}(a^x+b^x+c^x) . \]

- Fri Nov 04, 2016 8:23 pm
- Forum: Secondary Level
- Topic: An exercise
- Replies:
**3** - Views:
**2227**

### Re: An exercise

oops! I forgot about the case for n=1. :3

- Wed Nov 02, 2016 12:17 am
- Forum: Algebra
- Topic: Inequality with condition $xy+yz+zx=3xyz$
- Replies:
**2** - Views:
**2132**

### Re: Inequality with condition $xy+yz+zx=3xyz$

I think 2 is not annoying here as you can simply have by AM-GM,

\[ \sum_{cyc} (x^2y+\dfrac{1}{y}) \geq 2\sum_{cyc} x \]

the desired inequality!!!

\[ \sum_{cyc} (x^2y+\dfrac{1}{y}) \geq 2\sum_{cyc} x \]

the desired inequality!!!

- Tue Nov 01, 2016 7:51 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2016 Problem 6
- Replies:
**1** - Views:
**2517**

### Re: IMO 2016 Problem 6

The main idea of my solution Part (a) : WLOG, we may assume that the line segments are actually chords of a circle and all the chords intersect one another inside the circle. Now we shall label the endpoints one after another like $A_1, A_2,....A_{2n}$. Thus points $A_k$ and $A_{k+n}$ are the end p...

- Tue Nov 01, 2016 7:06 pm
- Forum: Secondary Level
- Topic: An exercise
- Replies:
**3** - Views:
**2227**

### An exercise

Let $a_1,a_2,......a_n$ are positive real numbers such that $\sum_{i=1}^{n} \dfrac{1}{a_i} = 1$. Prove that,

\[ \sum_{i=1}^{n} \dfrac{a_i^2}{i} > \dfrac{2n}{n+1} \]

\[ \sum_{i=1}^{n} \dfrac{a_i^2}{i} > \dfrac{2n}{n+1} \]

- Mon Oct 31, 2016 11:44 pm
- Forum: Social Lounge
- Topic: Chat thread
- Replies:
**53** - Views:
**39044**

### Re: Chat thread

well, I am Mehedi Hasan from Chittagong. & I desperately love math & programming ^_^

- Sat Oct 22, 2016 11:20 pm
- Forum: Algebra
- Topic: Minimum & maximum
- Replies:
**1** - Views:
**1869**

### Re: Minimum & maximum

My solution The given equation is equivalent to, $p+r= \dfrac{2015-r}{q} $ or, $p+q+r= \dfrac{2015-r}{q} + q$ Now let $a= \dfrac{2015-r}{q}$ and $b=q$. Then we have, \[ p+q+r=a+b \] with $q=b$, $r=2015-ab$ and $p=a+ab-2015$. So now we've to actually find the minimum and maximum values of $a+b$. As, ...