With x,y,z \(\in R^+\) that satisfies \(x\ge z\). 

Show that: \(\dfrac{xz}{y^2+yz}+\dfrac{yz}{xz+yz}+\dfrac{x+2z}{x+z}\ge\dfrac{5}{2}\)

With a,b,c are three non-negative numbers that satisfy a + b + c = 3

Show that: \(\Sigma\left(a\sqrt{a+8}+2\right)\ge15\)

We denote that: \(A=x^2+x-24\)

<=> \(4A=4x^2+4x-96\)

<=> \(4A=\left(2x+1\right)^2-97\)

<=> \(4A=\left(2x+1-\sqrt{97}\right)\left(2x+1+\sqrt{97}\right)\)

<=> \(A=\dfrac{\left(2x+1-\sqrt{97}\right)\left(2x+1+\sqrt{97}\right)}{4}\)

<=> \(A=\left(x+\dfrac{1}{2}-\dfrac{\sqrt{97}}{2}\right)\left(x+\dfrac{1}{2}+\dfrac{\sqrt{97}}{2}\right)\)

@8#3@ = (8 + 8) x (6 - 3) = 16 x 3 = 48

Let a,b,c > 0 that satify a + b + c = 1

Find minimum value of  \(P=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}+\dfrac{1}{9abc}\)

We have : \(2^{18}=262144=262121+23\)

Total time we can calculate since John was born until he was married is : 

\(S=2^{18}\left(h\right)+1\left(h\right)=262144\left(h\right)+1\left(h\right)=262128\left(h\right)+17\left(h\right)\)  

    \(=10920\left(days\right)+17\left(h\right)=10922\left(days\right)+2\left(days\right)+17\left(h\right)\)

    \(=1560\left(weeks\right)+2\left(days\right)+17\left(h\right)\)

Because after a week , the day that Fransisco was born was unchange so it still on Tuesday 

After Tuesday 2 days is Thurday 

So , exactly the day that Francisco was married is at 17:00 on Thursday 

Let a,b,c are positive real numbers satisfy \(a^2+b^2+c^2=1\)

Show that : \(\Sigma\dfrac{a^3}{b+c}\ge\dfrac{1}{2}\)

\(\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{3}\right)^3\)

<=> \(8.\dfrac{a^3+b^3}{2}\ge a^3+3a^2b+3ab^2+b^3\)

<=> \(3\left(a^3+b^3\right)\ge3a^2b+3ab^2\)

<=> \(a^3+b^3-a^2b-ab^3\ge0\)

<=> \(\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)

<=> \(\left(a+b\right)\left(a-b\right)^2\ge0\)  (Right)

So ...... 

Applying Cauchy's inequality for 4 numbers , we have 

\(a^4+b^4+c^4+d^4\ge4.\sqrt[4]{a^4.b^4.c^4.d^4}=4abcd\)

Equation occur <=> a = b = c = d 

Applying Cauchy's inequality , we have 

\(\left(a^4+1\right)+\left(b^4+1\right)\ge2a^2+2b^2=2\left(a^2+b^2\right)\ge2.2ab=4ab\)

Equation occur <=> a = b = \(\pm1\)

\(a^4+b^4\ge a^3b+ab^3\)

<=> \(a^4+b^4-a^3b-ab^3\ge0\)

<=> \(a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

<=> \(\left(a^3-b^3\right)\left(a-b\right)\ge0\)

<=> \(\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Cause \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\in R\\a^2+ab+b^2=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\end{matrix}\right.\)

So ...... 

You have to have a condition : \(a,b>0\)

a) \(\left(a+b\right)^2\ge4ab\)

<=> \(a^2+2ab+b^2\ge4ab\)

<=>  \(a^2-2ab+b^2\ge0\)

<=>  \(\left(a-b\right)^2\ge0\)  (It's true)

=> \(\left(a+b\right)^2\ge4ab\)

b) Applying Cauchy's inequality , we have

\(a+b\ge2\sqrt{ab}\)

Equation occur <=> a = b 

Let P is an integer number , show that :

\(GCD\left(P^2-1;P^2+1\right)\ne1\)

We have 

\(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{b}{c}\Leftrightarrow\dfrac{10a+b}{10b+c}=\dfrac{b}{c}=\dfrac{10a+b-b}{10b+c-c}=\dfrac{10a}{10b}=\dfrac{a}{b}\)

=> \(\dfrac{a}{b}=\dfrac{b}{c}\)

1^st way : We have

\(\sqrt{8}^2=8=6+2\)

\(\left(\sqrt{5}+1\right)^2=5+2\sqrt{5}+1=6+2\sqrt{5}\)

Cause \(2\sqrt{5}\ge2\sqrt{4}=4>2\)

So \(\sqrt{8}< \sqrt{5}+1\)

2^st way :

\(\sqrt{8}< \sqrt{9}=3=2+1=\sqrt{4}+1< \sqrt{5}+1\)

So ...... 

Give a² + b² = c² + d² = 2017 and ac + bd = 0 

Calculate the value of A = ab + cd 

\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{2^2}\)

<=> \(4\left(a^2+b^2\right)\ge2\left(a+b\right)^2\)

<=> \(4\left(a^2+b^2\right)\ge2\left(a^2+2ab+b^2\right)\)

<=>  \(\left(2a^2+2b^2-4ab\right)\ge0\)

<=> \(2\left(a-b\right)^2\ge0\)   (Right)

So .... 

\(\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}=\dfrac{ac}{2+ac+2c}+\dfrac{ac.b}{abc.c+abc+ac}+\dfrac{2c}{ac+2c+2}\)

                                                                          \(=\dfrac{ac}{2+ac+2c}+\dfrac{2}{2c+2+ac}+\dfrac{2c}{ac+2c+2}=1\)

+) With n = 1 then 

\(3^{2n+3}+40n-27=256⋮64\)

+) Suppose that \(\left(3^{2n+3}+40n-27\right)⋮64\)

And we will prove \(\left(3^{2\left(n+1\right)+3}+40\left(n+1\right)-27\right)⋮64\) , too 

Also \(3^{2\left(n+1\right)+3}+40\left(n+1\right)-27=3^{2n+3}.9+40n+40-27\)

                                                         \(=3^{2n+3}.9+9.40n-27.9-8.40n+27.8+40\)

                                                         \(=9\left(3^{2n+3}+40n-27\right)-320n+256\)

                                                         \(=9\left(....\right)-64\left(5n+4\right)⋮64\)

So \(\left(3^{2\left(n+1\right)+3}+40\left(n+1\right)-27\right)⋮64\)  is true 

Conclude \(\left(3^{2n+3}+40n-27\right)⋮64\)        \(\forall n\in N\)

+) With n = 1 then 

\(16^n-15n-1=16-15-1=0⋮225\)

+) With n = 2 then 

\(16^n-15n-1=16^2-15.2-1=225⋮225\)

Suppose \(\left(16^n-15n-1\right)⋮225\)

We are going to prove that \(\left(16^{n+1}-15\left(n+1\right)-1\right)⋮225\) too

We can see \(16^{n+1}-15\left(n+1\right)-1=16.16^n-15n-15-1=\left(16^n-15n-1\right)+15.16^n-15\)

Following Inductive hypothesis , \(\left(16^n-15n-1\right)⋮225\)

And \(15.16^k-15=15\left(16^k-1\right)⋮15.15=225\)

So \(\left(16^{n+1}-15\left(n+1\right)-1\right)⋮225\)

Conclude \(\left(16^n-15n-1\right)⋮225\) \(\forall n\in N\)

\(34^2+66^2+68.66=\left(34^2+66^2+2.34.66\right)=\left(34+66\right)^2=100^2=10000\)

Find the triple pairs of integer (x;y;z) that satisfy: 

\(2^x+2^y+2^z=2336\)

\(55^{n+1}-55^n=55^n\left(55-1\right)=55^n.54⋮54\)

So .... 

Another way :

Applying Cauchy's inequality , we have 

\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{ab}{2\sqrt{ab}}+\dfrac{bc}{2\sqrt{bc}}+\dfrac{ca}{2\sqrt{ca}}=\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\)

We have an extra inquality : With a,b,c are positive then \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Prove: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

<=>  \(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\)  (Right) 

So \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)  is true

We have 

\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\le\dfrac{1}{2}.\left(a+b+c\right)\)

Equation occur <=> a = b = c 

a) With x = 2008 => \(A=2\cdot\left|2008-2007\right|+\left|2008-2009\right|=2+1=3\)

b) Cause A = \(2\cdot\left|x-2007\right|+\left|x-2009\right|\)

So for A = x - 2010

<=> \(2\cdot\left|x-2007\right|+\left|x-2009\right|=x-2010\)   (*)

+) With x < 2007 

(*) <=> 2(2007 - x) + (2009 - x) = x - 2010 

     <=>  6023 - 3x = x - 2010

     <=>  6023 + 2010 = 4x

     <=>  x = \(\dfrac{8033}{4}>2007\)  (Removed) 

+) With \(2007\le x\le2009\)

(*) <=> \(2\cdot\left(x-2007\right)+\left(2009-x\right)=x-2010\)

     <=> \(x-2005=x-2010\) (impossible) 

+) With \(x>2009\)

(*)  <=> \(2\cdot\left(x-2007\right)+\left(x-2009\right)=x-2010\)

      <=>  \(3x-6023=x-2010\)

      <=> \(2x=4013\)

      <=>  \(2x=\dfrac{4013}{2}< 2009\)  (Removed)

So , There isn't any roots of this equation 

c) Applying this inequality 

With a,b are two numbers , we have \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Equation occur <=> ab \(\ge\) 0

So \(A=2\cdot\left|x-2007\right|+\left|x-2009\right|\)

         \(=\left|x-2007\right|+\left|2009-x\right|+\left|x-2007\right|\ge\left|x-2007+2009-x\right|+\left|x-2007\right|\)

                                                                                  \(=2+\left|x-2007\right|\ge2\)

=> Min(A) = 2 <=> \(\left\{{}\begin{matrix}x=2007\\\left(x-2007\right)\left(2009-x\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2007\\2007\le x\le2009\end{matrix}\right.\)

So Min(A) = 2 <=> x = 2007 

Applying Cauchy's inequaility , we have 

\(\dfrac{ab}{a+b}\le\dfrac{\dfrac{\left(a+b\right)^2}{4}}{a+b}=\dfrac{a+b}{4}\)

\(\dfrac{bc}{b+c}\le\dfrac{\dfrac{\left(b+c\right)^2}{4}}{b+c}=\dfrac{b+c}{4}\)

\(\dfrac{ca}{c+a}\le\dfrac{\dfrac{\left(c+a\right)^2}{4}}{c+a}=\dfrac{c+a}{4}\)

So \(\sum\left(\dfrac{ab}{a+b}\right)\le\dfrac{a+b+b+c+c+a}{4}=\dfrac{2\left(a+b+c\right)}{4}=\dfrac{1}{2}\left(a+b+c\right)\)

Equation occur

<=> a = b = c 

Give an equilateral hexagon. Six birds land on six tops. After a few minutes, six bird fly away and land on 6 tops again. (They don's have to land on the top that they have landed). Show that , there are exist at least three birds that the triangle they have made before fly equals to the triangle they have made after land.

Cause \(a^4,b^4,c^4,d^4\) are square numbers 

So when they divided for 3 , Their remainder must be 1 or 0

<=> \(\left(a^4+b^4+c^4+d^4\right)\equiv1\left(mod3\right)\)

or \(\left(a^4+b^4+c^4+d^4\right)\equiv0\left(mod3\right)\)

But \(2018\equiv2\left(mod3\right)\)

So there isn't any root of (a;b;c;d) that satisfy the equation

It also means that we can't calculate the value of \(a^2+b^2+c^2+d^2\)

1. 12 = \(6+5+\dfrac{3}{3}\)

2. 12 = \(3+3+6\)

3. 12 = \(\sum\limits^6_3\left(X\right)-6\)

4. 12 = \(\prod\limits^6_5\left(X\right)-3\cdot5-3\)

\(4^{50}=2^{100}=2^{10}.2^{10}=\left(...4\right).\left(...4\right)=...6\)

So ....... 

How about using Mincopsky's Inequality ? 

Suppose p is an odd prime number and   \(m=\dfrac{9p-1}{8}\)

Prove that : m is an odd integer not divisible by 3 and 

                      \(3^{m-1}\equiv1\)   (mod m)

With a,b,c are three lengths of three edges of a triangle.

Show that : \(A=\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}\ge3\)