Compare the numbers $X=(\log_2(\sqrt{5}+1))^3$ and $Y=1+\log_2(\sqrt{5}+2)$.

First, note that $5\gt 1$, which gives $2\sqrt{5}\gt 2$ and further translates into $5+2\sqrt{5}+1=(\sqrt{5}+1)^2\gt 8$, which implies $\sqrt{5}+1\gt 2^{\frac{3}{2}}$, taking base 2 logarithm of both sides of the inequality we get:

A collection of intriguing competition level problems for secondary school students.

Friday, April 29, 2016

Wednesday, April 27, 2016

### Find, in terms of $a$, where $a \gt 0$, the minimum value of the expression [MATH]\frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}[/MATH] for all non-negative real $x,\,y$ and $z$ such that $x+y+z=1$.

Find, in terms of $a$, where $a \gt 0$, the minimum value of the expression [MATH]\frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}[/MATH] for all non-negative real $x,\,y$ and $z$ such that $x+y+z=1$.

Since $9xyz≥4(xy+yz+zx)-1$ (by the Schur's inequality), we can transform the objective function as

Since $9xyz≥4(xy+yz+zx)-1$ (by the Schur's inequality), we can transform the objective function as

Friday, April 22, 2016

### What is the numerical value of the expression [MATH]\frac{(a + b)(b + c)(c + a)(8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8))}{abc}[/MATH]?

Let $a,\,b,\,c \in \Bbb{R}$ such that [MATH]\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}[/MATH]

What is the numerical value of the expression [MATH]\frac{(a + b)(b + c)(c + a)(8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8))}{abc}[/MATH]?

My solution:

What is the numerical value of the expression [MATH]\frac{(a + b)(b + c)(c + a)(8^{100}-7(8^{99})-7(8^{98})-\cdots-7(8))}{abc}[/MATH]?

My solution:

Tuesday, April 19, 2016

### Let $a,\,b$ and $c$ be positive real numbers satisfying $a+b+c=1$. Prove that $9abc\ge7(ab+bc+ca)-2$.

Let $a,\,b$ and $c$ be positive real numbers satisfying $a+b+c=1$.

Prove that $9abc\ge7(ab+bc+ca)-2$.

Prove that $9abc\ge7(ab+bc+ca)-2$.

Sunday, April 17, 2016

### For real numbers [MATH]0\lt x\lt \frac{\pi}{2}[/MATH], prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$. (Second Solution)

For real numbers [MATH]0\lt x\lt \frac{\pi}{2}[/MATH], prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$.

My solution:

My solution:

Saturday, April 16, 2016

### For real numbers [MATH]0\lt x\lt \frac{\pi}{2}[/MATH], prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$. (First Solution)

For real numbers [MATH]0\lt x\lt \frac{\pi}{2}[/MATH], prove that $\cos^2 x \cot x+\sin^2 x \tan x\ge 1$.

MarkFL's solution:

MarkFL's solution:

Thursday, April 14, 2016

### Simplify [MATH]\tiny\frac{(2^{2^0}+3^{2^0})(2^{2^1}+3^{2^1})(2^{2^2}+3^{2^2})\cdots(2^{2^{10}}+3^{2^{10}})+2^{2048}}{3^{2048}}[/MATH].

Simplify [MATH]\frac{(2^{2^0}+3^{2^0})(2^{2^1}+3^{2^1})(2^{2^2}+3^{2^2})\cdots(2^{2^{10}}+3^{2^{10}})+2^{2048}}{3^{2048}}[/MATH].

My solution:

My solution:

Tuesday, April 12, 2016

### Let $a,\,b$ and $c$ be positive real that is greater than $1$ such that [MATH]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2[/MATH]. Prove that $\sqrt{a+b+c}\ge \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}$.

Let $a,\,b$ and $c$ be positive real that is greater than $1$ such that [MATH]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2[/MATH].

Prove that $\sqrt{a+b+c}\ge \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}$.

My solution:

Prove that $\sqrt{a+b+c}\ge \sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}$.

My solution:

Friday, April 8, 2016

### Prove, with no knowledge of the decimal value of $\pi$ should be assumed or used that [MATH]1\lt \int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx \lt \frac{2\sqrt{3}}{3}[/MATH].

Prove, with no knowledge of the decimal value of $\pi$ should be assumed or used that [MATH]1\lt \int_{3}^{5} \frac{1}{\sqrt{-x^2+8x-12}}\,dx \lt \frac{2\sqrt{3}}{3}[/MATH].

The solution below is provided by MarkFL:

We are given to prove:

The solution below is provided by MarkFL:

We are given to prove:

Wednesday, April 6, 2016

### Let a,b and c be positive real numbers with $abc = 1$, prove that [MATH]\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge 1[/MATH]

Let a,b and c be positive real numbers with $abc = 1$, prove that

[MATH]\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge 1[/MATH]

In the problem 4 as shown in quiz 22, I asked if you could approach the problem using the HÃ¶lder's inequality, I hope you have tried it before checking out with my solution to see why the HÃ¶lder's inequality wouldn't help:

[MATH]\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge 1[/MATH]

In the problem 4 as shown in quiz 22, I asked if you could approach the problem using the HÃ¶lder's inequality, I hope you have tried it before checking out with my solution to see why the HÃ¶lder's inequality wouldn't help:

Monday, April 4, 2016

### Analysis Quiz 22: Proving An Inequality

Let a,b and c be positive real numbers with $abc = 1$, prove that

[MATH]\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge 1[/MATH]

Would you see turning the RHS of the inequality of 1 as $abc$ help?

[MATH]\frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge 1[/MATH]

**Question 1:**Would you see turning the RHS of the inequality of 1 as $abc$ help?

Saturday, April 2, 2016

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