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Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition: $a+x=b+y=c+z=1$ Prove the inequality [MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3[/MATH].

Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition:

$a+x=b+y=c+z=1$

Prove the inequality [MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3[/MATH].

My solution:

Rewrite the intended LHS of the inequality strictly in terms of a, b and c, we have:

$x=1-a,\,y=1-b,\,z=1-c$

[MATH]\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)[/MATH]

[MATH]=\left(abc+(1-a)(1-b)(1-c)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\left(abc+1+ab+bc+ca-(a+b+c)-abc\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\left(1+ab+bc+ca-(a+b+c)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\left(1-a(1-b)-b(1-c)-c(1-a)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]=\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}-1-a(1-b)\left(\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)[/MATH]

[MATH]\,\,\,\,\,\,-1-b(1-c)\left(\frac{1}{a(1-b)}+\frac{1}{c(1-a)}\right)-1-c(1-a)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}\right)[/MATH]

[MATH]=\frac{1-b(1-c)-c(1-a)}{a(1-b)}+\frac{1-a(1-b)-c(1-a)}{b(1-c)}+\frac{1-a(1-b)-b(1-c)}{c(1-a)}-3[/MATH]

[MATH]=\frac{1-b+bc-c+ca}{a(1-b)}+\frac{1-a+ab-c+ac}{b(1-c)}+\frac{1-a+ab-b+bc}{c(1-a)}-3[/MATH]

[MATH]=\frac{1-b-c(1-b)+ca}{a(1-b)}+\frac{1-c-a(1-c)+ab}{b(1-c)}+\frac{1-a-b(1-a)+bc}{c(1-a)}-3[/MATH]

[MATH]=\frac{(1-b)(1-c)+ca}{a(1-b)}+\frac{(1-a)(1-c)+ab}{b(1-c)}+\frac{(1-b)(1-a)+bc}{c(1-a)}-3[/MATH]

[MATH]=\frac{1-c}{a}+\frac{c}{1-b}+\frac{1-a}{b}+\frac{a}{1-c}+\frac{1-b}{c}+\frac{b}{1-a}-3[/MATH]

[MATH]\ge 6\sqrt[6]{\frac{1-c}{a}\cdot \frac{c}{1-b}\cdot\frac{1-a}{b}\cdot\frac{a}{1-c}\cdot\frac{1-b}{c}\cdot\frac{b}{1-a}}-3[/MATH] (By the AM-GM inequality, with $1-a,\,1-b,\,1-c$ are all positive)

[MATH]= 6-3[/MATH]

[MATH]= 3[/MATH] (Q.E.D.)

Evaluate [MATH]\small\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor[/MATH] without using a calculator.

Evaluate [MATH]\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor[/MATH] without using a calculator.

My solution:

[MATH]\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)[/MATH]

[MATH]=\frac{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}{\left(-\sqrt{2}-\sqrt{3}-\sqrt{6}\right)}[/MATH]

[MATH]=\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}[/MATH]

By the Cauchy-Schwarz inequality, we have:

[MATH]\begin{align*}\sqrt{2}+\sqrt{3}+\sqrt{6}&<\sqrt{1+1+1}\sqrt{2+3+6}\\&=\sqrt{33}\end{align*}[/MATH]

Hence [MATH]\frac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}>\frac{23}{\sqrt{33}}[/MATH].

From $528\lt 529$ we get, after taking the square root on both sides and rearranging:

$4\lt \dfrac{23}{\sqrt{33}}$

$\therefore \dfrac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\gt \dfrac{23}{\sqrt{33}}\gt 4$

On the other hand,

From $50\gt 49$, we get:

$\sqrt{2}\gt \dfrac{7}{5}$

From $12\gt 9$, we get:

$\sqrt{3}\gt \dfrac{3}{2}$

From $6\gt 4$, we get:

$\sqrt{6}\gt 2$

Adding them up gives:

$\sqrt{2}+\sqrt{3}+\sqrt{6}\gt 4.9$

$\therefore \dfrac{23}{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)}\lt \dfrac{23}{4.9}=4.69$.

We can conclude by now that [MATH]\left\lfloor{\left(-\sqrt{2}+\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)}\right\rfloor=4.[/MATH]

If $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, prove [MATH]\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}[/MATH].

Show that if $a,\,b$ and $c$ are the lengths of the sides of a right triangle with hypotenuse $c$, then
(a) [MATH]\frac{c}{\sqrt{ab}}\ge \sqrt{2}[/MATH]
(b) [MATH]\frac{(c − a)(c − b)}{(c + a)(c + b)}\le 17 − 12\sqrt{2}[/MATH].

My solution:

(a)

$\begin{align*}c^2&=a²+b²\\&≥2ab \,\,\,\text{(by the AM-GM inequality)}\end{align*}$

$\dfrac{c^2}{ab}\ge 2$
Taking the square root on both sides completes the proof, i.e. $\dfrac{c}{\sqrt{ab}}\ge \sqrt{2}$, equality occurs when $a=b$.

(b)

We have $c^2−a^2=b^2 \implies c−a=\dfrac{b^2}{c+a}$.

By the similar token, we also have $c−b=\dfrac{a^2}{c+b}$, if we're going to replace these two into the original LHS of the inequality, we get:

[MATH]\begin{align*}\frac{(c − a)(c − b)}{(c + a)(c + b)}&=\frac{(b^2)(a^2)}{(c + a)^2(c + b)^2}\\&=\left(\frac{ab}{(c + a)(c + b)}\right)^2\\&=\left(\frac{ab}{c^2+c(a+b)+ab}\right)^2\\&=\left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(a+b)}{ab}+1}\right)^2\\& \le \left(\frac{1}{\dfrac{c^2}{ab}+\dfrac{c(2\sqrt{ab})}{ab}+1}\right)^2\,\,\text{(by the AM-GM inequality)}\\& = \left(\frac{1}{\left(\dfrac{c}{\sqrt{ab}}\right)^2+\dfrac{2c}{\sqrt{ab}}+1}\right)^2\\& \le \left(\frac{1}{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\right)^2\\&= \left(3-2\sqrt{2}\right)^2\\&= 17-12\sqrt{2}\,\,\,\text{Q.E.D.}\end{align*}[/MATH]

Equality occurs when $a=b$.

Solve for real solution of the system below: [MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH].

Solve for real solution of the system below:

[MATH]\left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1[/MATH].

My solution:

First, observe that the LHS of the equality must yield an integer, this tells us the fractional part of $x$ must be a zero, so this turns the whole equation as:

[MATH]x^3+x^2+x=-1\\x^3+x^2+x+1=0\\(x+1)(x^2+1)=0[/MATH]

This implies $x=-1$ is the only real solution to the system.

Solve for real solutions for the system: $x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$

Solve for real solutions for the system:

$x+y+z=-a,\\x^2+y^2+z^2=a^2,\\x^3+y^3+z^3=-a^3.$

My solution:

Let $x,\,y$ and $z$ be the real roots for a cubic polynomial.

From the relation $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, we have:

$(-a)^2=a^2+2(xy+yz+zx)$

$a^2=a^2+2(xy+yz+zx)\implies xy+yz+zx=0$

From the relation $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$, we have:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$

$-a^3-3xyz=(-a)(a^2-0)$

$-a^3-3xyz=-a^3\implies xyz=0$

We can now form the cubic polynomial in $t$ where its roots are $x,\,y$ and $z$:

$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz=0$

$t^3-(-a)t^2+(0)t-0=0$

$t^2(t+a)=0$

Obviously $t=0$ is the repeated root and the other root is $t=-a$.

Therefore we get the solution:

$(x,\,y,\,z)=(0,\,0,\,-a),\,(0,\,-a,\,0),\,(-a,\,0,\,0)$

Show that [MATH]abc\le (ab+bc+ca)(a^2+b^2+c^2)^2[/MATH] for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$

Show that [MATH]abc\le (ab+bc+ca)(a^2+b^2+c^2)^2[/MATH] for all positive real $a,\,b$ and $c$ such that $a+b+c=1.$

My solution:

[MATH](ab+bc+ca)(a^2+b^2+c^2)^2[/MATH]

[MATH]\ge \frac{(ab+bc+ca)(a^2+b^2+c^2)(a+b+c)^2}{3}[/MATH] since $3(a^2+b^2+c^2)\ge (a+b+c)^2$

[MATH]= \frac{(ab+bc+ca)(a^2+b^2+c^2)}{3}[/MATH] since $a+b+c=1$

[MATH]= \frac{a^3b+b^3c+ac^3+a^3c+ab^3+bc^3+a^2bc+ab^2c+abc^2}{3}[/MATH]

[MATH]= \frac{\left(\frac{a^2}{\frac{1}{ab}}+\frac{b^2}{\frac{1}{bc}}+\frac{c^2}{\frac{1}{ca}}\right)+\left(\frac{a^2}{\frac{1}{ac}}+\frac{b^2}{\frac{1}{ab}}+\frac{c^2}{\frac{1}{bc}}\right)+a^2bc+ab^2c+abc^2}{3}[/MATH]

[MATH]\ge \frac{\left(\frac{(a+b+c)^2}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\right)+\left(\frac{(a+b+c)^2}{\frac{1}{ac}+\frac{1}{ab}+\frac{1}{bc}}\right)+a^2bc+ab^2c+abc^2}{3}[/MATH] (By the Titu's Lemma)

[MATH]= \frac{\left(\frac{1}{\frac{a+b+c}{abc}}\right)+\left(\frac{1}{\frac{a+b+c}{abc}}\right)+a^2bc+ab^2c+abc^2}{3}[/MATH]

(since [MATH]a+b+c=1[/MATH] and [MATH]\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{c}{abc}+\frac{a}{abc}+\frac{b}{cab}=\frac{a+b+c}{abc})[/MATH]

[MATH]= \frac{\left(abc+abc\right)+abc(a+b+c)}{3}[/MATH]

[MATH]= \frac{3abc}{3}[/MATH]

[MATH]= abc[/MATH] (Q.E.D)

Solve for real solution for $(1+x^2)(1+x^3)(1+x^5)=8x^5$.

Solve for real solution for $(1+x^2)(1+x^3)(1+x^5)=8x^5$.

My solution:

For $x\lt 0$, we have a positive left hand side value and a negative right hand side value. So $x$ can never be a negative value.

For $x\gt 1$, we have:

$1+x^2\gt 2x,\,(1+x^3)(1+x^5)=1+x^3+x^5+x^8\gt 4x^4$ so $(1+x^2)(1+x^3)(1+x^5)\gt 8x^5$, which really is $8x^5\gt 8x^5$, which leads to a contradiction.

For $0\le x \le 1$:

$f(x)=(1+x^2)(1+x^3)(1+x^5)$ has its first derivative of $f'(x)\gt 0$ and so $f$ is an increasing function and so does $f(x)=8x^5$.

That means they can intersect at most once, and by inspection, it is not hard to see that $x=1$ is the only real solution to the system.

Find the real solution(s) to the system $\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$.

Find the real solution(s) to the system

$\sqrt{4a-b^2}=\sqrt{b+2}+\sqrt{4a^2+b}$.

My solution:

Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real $a,\,b$ and $c$.

Prove that [MATH]\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+b^3+abc}\le \frac{1}{abc}[/MATH] for all positive real $a,\,b$ and $c$.

My solution:

Let $a,\,b,\,c,\,x,\,y$ and $z$ be strictly positive real numbers, prove that [MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH].

Let $a,\,b,\,c,\,x,\,y$ and $z$ be strictly positive real numbers, prove that

[MATH](a+x)(b+y)(c+z)+4\left(\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}\right)\ge 20[/MATH].

Given that [MATH]\frac{\sin 3x}{\sin x}=\frac{6}{5}[/MATH], what is the ratio of [MATH]\frac{\sin 5x}{\sin x}[/MATH]?

Given that [MATH]\frac{\sin 3x}{\sin x}=\frac{6}{5}[/MATH], what is the ratio of [MATH]\frac{\sin 5x}{\sin x}[/MATH]?

My solution:

For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

Hello readers!

In my previous blog post, I asked the readers to spot the factual mistake(s) that I might have or might not have made in the solution (of mine) to one delicious inequality problem.

Today, I am going to discuss with you the mistake that I intentionally made.

For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

For positive reals $a,\,b,\,c$, prove that [MATH]\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\gt 2[/MATH].

Hello all!

Today I'm going to post something that is going to be very different than my style in my previous blog posts, as today I wanted to train students to spot the factual mistake(s) that I might have or might not have made in the following solution (of mine) to today's delicious inequality problem.

Prove the following inequality holds: $\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.

Let the reals $a, b, c∈(1,\,∞)$ with $a + b + c = 9$.

Prove the following inequality holds:

$\sqrt{(\log_3a^b +\log_3a^c)}+\sqrt{(\log_3b^c +\log_3b^a)}+\sqrt{(\log_3c^a +\log_3c^b)}\le 3\sqrt{6}$.

My solution:

Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.

Let the real $x\in \left(0,\,\dfrac{\pi}{2}\right)$, prove that $\dfrac{\sin^3 x}{5}+\dfrac{\cos^3 x}{12}≥ \dfrac{1}{13}$.

Prove that $x^2 + y^2+ z^2\le xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

Prove that $x^2 + y^2+ z^2\le  xyz + 2$ where the reals $x,\,y,\, z\in [0,1]$.

For all $x,\,y,\, z\in [0,1]$, we know $x^2 + y^2+ z^2\le x+y+z$.

If one root of $4x^2+2x-1= 0 $ be $\alpha$, please show that other root is $4\alpha^3-3\alpha$.

If one root of $4x^2+2x-1= 0 $ be $\alpha$, please show that other root is $4\alpha^3-3\alpha$.

My solution:

The relation of $2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1$.

Prove that if in a triangle $ABC$ we have the following equality that holds

$2\cos A \cos B \cos C + \cos A \cos B + \cos B \cos C + \cos C \cos A = 1$

then the triangle will be an equilateral triangle.

In any triangle $ABC$, we have the following equality that holds:

Compare which of the following is bigger: [MATH]1016^{11}\cdot 3016^{31}[/MATH] versus [MATH]2016^{42}[/MATH]

Compare which of the following is bigger:

[MATH]1016^{11}\cdot 3016^{31}[/MATH] versus [MATH]2016^{42}[/MATH]

My solution:

Prove that : [MATH]\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}[/MATH] for all positive reals $a$ and $b$.

Prove that :
[MATH]\frac{\sqrt{a^2+b^2}}{a+b}+\sqrt{\frac{ab}{a^2+b^2}}\le \sqrt{2}[/MATH] for all positive reals $a$ and $b$.

My solution:

Step 1:

Solve for real solution(s) for [MATH]x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)[/MATH].

Solve for real solution(s) for [MATH]x^2− x + 1 = (x^2+ x + 1)(x^2+ 2x + 4)[/MATH].

My solution:

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 [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

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:

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 [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:

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:

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:

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, 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:

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 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

[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?

Quiz 22: Proving An Inequality

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

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 [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:

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:

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]

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:

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].

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

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!

Math Joke: Square Root

Prove that $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$.

Let $a,\,b,\,c,\,d$ be real numbers such that $a+b+c+d=0$.

Prove that $(a^3+b^3+c^3+d^3)^2=9(ab-cd)(bc-ad)(ca-bd)$.

My solution:

First, we draw some very useful and helpful identities from the given equality,  $a+b+c+d=0$:

Analysis Quiz 20: Multiple-Choice Algebra Test

Answer the following questions based on the evaluation of a sum (without the help of calculator) below:

Evaluate [MATH] \frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}[/MATH].

Quiz 20: Multiple-Choice Algebra Test

Slideshow 14: Train Students To Think Effectively & Make The Right Decision

Evaluate [MATH]\small \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)[/MATH].

Evaluate [MATH]\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)[/MATH].

My solution:

First, note that $6x^4-12x^3-x+2$ can be factorized as $6(2)^4-12(2)^3-(2)+2=96-96-2+2=0$, factorize it using the long polynomial division by the factor $x-2$, we get $6x^4-12x^3-x+2=(x-2)(6x^3-1)$.

Find all real solutions for the system [MATH]4x^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH] where [MATH]\left\lfloor{x}\right\rfloor[/MATH] represents the floor of $x$.

Find all real solutions for the system [MATH]4x^2-40\left\lfloor{x}\right\rfloor+51=0[/MATH] where [MATH]\left\lfloor{x}\right\rfloor[/MATH] represents the floor of $x$.

My solution:

First, notice that if we rewrite the equality as  [MATH]4x^2+51=40\left\lfloor{x}\right\rfloor[/MATH], we can tell [MATH]\left\lfloor{x}\right\rfloor[/MATH] must be a positive figure.

Find the number of real solutions for the system $x^4-x^3+x^2-4x-12=0$.

Find the number of real solutions for the system $x^4-x^3+x^2-4x-12=0$.

First, we let $f(x)=x^4-x^3+x^2-4x-12$. $f(x)$ clearly is a quartic function and it has at most 4 real roots.

If we can factorize $f(x)$ as $f(x)=(x-a)(x-b)(x-c)(x-d)$, then $f(x)$ has 4 real roots.

Analysis Quiz 19: Multiple-Choice Test (Improve Analytical Skill)

Question 1: If you're asked to simplify [MATH]\left(1+\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)^2[/MATH], do you think by turning the $1$ as [MATH]\left(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\right)[/MATH] is feasible in order to simplify the expression?

A. Yes.
B. No.

Quiz 19: Multiple-Choice Test (Improve Analytical Skill)

Factorize $5(a^3+b^3+c^3)-3(a^2+b^2+c^2)(a+b+c)+12abc$.

Factorize $5(a^3+b^3+c^3)-3(a^2+b^2+c^2)(a+b+c)+12abc$.

My solution:

The given expression is written so neatly and beautifully at first glance, it's like it couldn't be factored. And we feel so reluctant to expand the second product, but we have to, if we want to factor the expression correctly, we have to get a clearer picture of what this expression is all about by expanding and then rearranging terms in decreasing order:

Analysis Quiz 18: Multiple-Choice Test (Develop Problem Solving Skill)

Please answer the following questions based on the proving of the inequality below:
Given $a,\,b,\,c$ and $d$ are all positive real numbers such that $a+b+c+d=4$. Prove that [MATH]\sum_{\text{cyclic}}^{} \frac{a}{a^3+8}\le \frac{4}{9}[/MATH].

Question 1: Do you think we can stick to the approach we employed in the previous quiz (17) for proving this inequality?
A. Yes.
B. No.

Quiz 18: Multiple-Choice Test (Develop Problem Solving Skill)

IMO Inequality Problem

Let $a,\,b,\,c$ be real numbers greater than $2$ such that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$.

Prove that $(a-2)(b-2)(c-2)\le 1$.

My solution:

Note that

$(a−2)(b−2)(c−2)$

A system of the form $a+b\sqrt{2}$.

Given $x$ and $y$ are of the form $a+b\sqrt{2}$ ($a$ and $b$ are both positive integers) that satisfy the equation

$2x+y-\sqrt{3x^2+3xy+y^2}=2+\sqrt{2}$.

Find such $x$ and $y$.

Find the minimum value of $\tan^7 x(\tan y \tan z-1)+\tan^7 y(\tan x\tan z-1)+\tan^7 z(\tan x \tan y-1)$

$x$, $y$ and $z$ are three acute angles from a triangle.

Find the minimum value of $\tan^7 x(\tan y \tan z-1)+\tan^7 y(\tan x\tan z-1)+\tan^7 z(\tan x \tan y-1)$.

The trick for solving this problem fast and effective depends on if you could use the implicit relation between $x,\,y$ and $z$ when they are the angles from a triangle:

What else could we generate from $A+B+C=\pi$, when $A,\, B,\,C$ are three angles from a triangle?

Following previous blog post, we have shown that if

If $A,\, B,\,C$ are three angles from a triangle, i.e. $A+B+C=\pi$, then we should have known the following equality holds.

[MATH]\tan A+\tan B+\tan C=\tan A\tan B\tan C[/MATH]

On this blog post, we now try to generate another relation between $A,\,B$ and $C$ for cotangent functions:

What could educators do to motivate students to learn well in mathematics?

Mathematics is one of the most powerful tools to shape the way we think and see the world. But it's no secret that success in learning math is very much depends on maintaining a high level of motivation. Without motivation and a sense of emotional involvement, it's hard if not difficult to have the stamina to keep learning.

Analysis Quiz 17: Multiple-Choice Test (Improve Logical Thinking and Develop Discerning Patterns Skills)

Question 1: Given $a,\,b,\,c$ and $d$ are all positive real numbers such that $a+b+c+d=4$. Prove that [MATH]\sum_{cyclic} \frac{a}{a^3+8}\lt \frac{12}{25}[/MATH].

Do you think you would need the given equality to prove for the inequality?

A. Yes.
B. No.

Answer:

Of course we need the given equality to prove for the target expression such that it must be less than [MATH]\frac{12}{25}[/MATH].

Quiz 17: Multiple-Choice Test (Improve Logical Thinking and Develop Discerning Patterns Skills)

Happy New Year 2016, everyone! :D