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

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:

Find, in terms of $a$, where $a \gt 0$, the minimum value of the expression $\displaystyle \frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}$ 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 $\displaystyle \frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}$ 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

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

Let $a,\,b,\,c \in \Bbb{R}$ such that $\displaystyle \frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}$

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

My solution:

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

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

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

My solution:

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

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

MarkFL's solution:

Simplify $\displaystyle \tiny\frac{(2^{2^0}+3^{2^0})(2^{2^1}+3^{2^1})(2^{2^2}+3^{2^2})\cdots(2^{2^{10}}+3^{2^{10}})+2^{2048}}{3^{2048}}$.

Simplify $\displaystyle \frac{(2^{2^0}+3^{2^0})(2^{2^1}+3^{2^1})(2^{2^2}+3^{2^2})\cdots(2^{2^{10}}+3^{2^{10}})+2^{2048}}{3^{2048}}$.

My solution:

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

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

My solution:

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

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

The solution below is provided by MarkFL:

We are given to prove:

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

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

$\displaystyle \frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge  1$

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

Analysis Quiz 22: Proving An Inequality

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

$\displaystyle \frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge  1$

Question 1:

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

Quiz 22: Proving An Inequality