Thursday, March 24, 2016

For $n\in N,n\geq 2$, prove that [MATH]\sum_{k=1}^{n}(\dfrac{1}{2k-1}-\dfrac{1}{2k})>\dfrac {2n}{3n+1}[/MATH].

For $n\in N,n\geq 2$, prove that
[MATH]\sum_{k=1}^{n}(\dfrac{1}{2k-1}-\dfrac{1}{2k})>\dfrac {2n}{3n+1}[/MATH].

Note that

Tuesday, March 22, 2016

Factorize $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$.

Factorize $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$.

My solution:

In a problem such as this one, two inclinations may arise:

Sunday, March 13, 2016

Prove $17^{14}\gt 31^{11}$

Prove $17^{14}\gt 31^{11}$.

My solution:

Thursday, March 10, 2016

Second Attempt: Let $a,\,b$ and $c$ be real numbers such that $(a-b)^3+(b-c)^3+(c-a)^3=9$. Prove that [MATH]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\ge \sqrt[3]{3}[/MATH].

Let $a,\,b$ and $c$ be real numbers such that $(a-b)^3+(b-c)^3+(c-a)^3=9$.

Prove that [MATH]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\ge \sqrt[3]{3}[/MATH].

Second attempt:

Tuesday, March 8, 2016

Let $a,\,b$ and $c$ be real numbers such that $(a-b)^3+(b-c)^3+(c-a)^3=9$. Prove that [MATH]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\ge \sqrt[3]{3}[/MATH].

Let $a,\,b$ and $c$ be real numbers such that $(a-b)^3+(b-c)^3+(c-a)^3=9$.

Prove that [MATH]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\ge \sqrt[3]{3}[/MATH].

My solution:

Sunday, March 6, 2016

Let $a\,b$ and $c$ be the sides of a triangle. Prove that [MATH]\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3[/MATH].

Let $a\,b$ and $c$ be the sides of a triangle.

Prove that [MATH]\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3[/MATH].

My solution:

[MATH]\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}[/MATH]

[MATH]=\frac{a^2}{a(b+c)-a^2}+\frac{b^2}{b(a+c)-b^2}+\frac{c^2}{c(a+b)-c^2}[/MATH]

[MATH]\ge 4\left(\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\right)\,\,\text{since}\,\,a^2+\frac{(b+c)^2}{4}\ge a(b+c)[/MATH]

[MATH]\ge 4\left(\frac{\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2}{3}\right)\,\,\text{since}\,\,3(x^2+y^2+z^2)\ge (x+y+z)^2[/MATH]

[MATH]\ge 4\left(\frac{\left(\frac{3}{2}\right)^2}{3}\right)\,\,\text{from the Nesbitt's inequality}

[MATH]\ge 3[/MATH]


Saturday, March 5, 2016

Evaluate [MATH]\left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor[/MATH].

Evaluate [MATH]\left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor[/MATH].

Let $x=2015$.

Therefore we see that we have:

Friday, March 4, 2016

Analysis Quiz 21: Multiple-Choice Test (Improve Logical And Common Sense Reasoning In Learning Mathematics)

Question 1: What can you conclude to the sum of the series based on the sequence listed as follows?
[MATH]P=\frac{110}{100+1}+\frac{110}{100+2}+\frac{110}{100+2}+\cdots+\frac{110}{100+10}[/MATH]

A. [MATH]P\gt 1[/MATH].
B. [MATH]P\gt 10[/MATH].
C. [MATH]P\gt 100[/MATH].

Thursday, March 3, 2016

Quiz 21: Multiple-Choice Test (Improve Logical And Common Sense Reasoning In Learning Mathematics)

Wednesday, March 2, 2016

Solve for real solutions of the system below: [MATH]\small \frac{1}{x-1}+\frac{2}{x-2}+\frac{6}{x-6}+\frac{7}{x-7}=x^2-4x-4[/MATH]

Solve for real solutions of the system below:

[MATH]\frac{1}{x-1}+\frac{2}{x-2}+\frac{6}{x-6}+\frac{7}{x-7}=x^2-4x-4[/MATH]

My solution:

First of all, notice that if one wants to solve the given system by clearing the fraction on the left, one would definitely end up with having to solve the polynomial of degree 6, which is a giant PITA!

Tuesday, March 1, 2016

Math Joke: Square Root