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

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

Note that

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:

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

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

My solution:

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

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

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

Second attempt:

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

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

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

My solution:

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

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

Prove that $\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3$.

My solution:

$\displaystyle \frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}$

$\displaystyle =\frac{a^2}{a(b+c)-a^2}+\frac{b^2}{b(a+c)-b^2}+\frac{c^2}{c(a+b)-c^2}$

$\displaystyle \ge 4\left(\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\right)\,\,\text{since}\,\,a^2+\frac{(b+c)^2}{4}\ge a(b+c)$

$\displaystyle \ge 4\left(\frac{\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2}{3}\right)\,\,\text{since}\,\,3(x^2+y^2+z^2)\ge (x+y+z)^2$

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

Evaluate $\displaystyle \left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor$.

Evaluate $\displaystyle \left\lfloor{\frac{2014^3}{(2015)(2016)}+\frac{2016^3}{2014(2015)}}\right\rfloor$.

Let $x=2015$.

Therefore we see that we have:

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?
$\displaystyle P=\frac{110}{100+1}+\frac{110}{100+2}+\frac{110}{100+2}+\cdots+\frac{110}{100+10}$

A. $\displaystyle P\gt 1$.
B. $\displaystyle P\gt 10$.
C. $\displaystyle P\gt 100$.

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

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

Solve for real solutions of the system below:

$\displaystyle \frac{1}{x-1}+\frac{2}{x-2}+\frac{6}{x-6}+\frac{7}{x-7}=x^2-4x-4$

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!

Math Joke: Square Root