Denote A = |x - 1| + |x - 3|     (You can write or not)

We have :

A = |x - 1| + |x - 3|

<=> A = |x - 1| + |-x + 3|

Apply inequality |a| + |b| \(\ge\) |a + b| 

=> A = |x - 1| + |-x + 3| \(\ge\) |x - 1 - x + 3| = |2| = 2

<=> \(\left\{{}\begin{matrix}x-1\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\le3\end{matrix}\right.\Leftrightarrow1\le x\le3\)

Convert  \(\dfrac{a}{b}=\dfrac{a}{d}\) to \(\dfrac{a}{b}=\dfrac{c}{d}\)

We have :

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{2a}{2b}=\dfrac{3c}{3d}\)

Apply the same sequence properties , we got :

\(\dfrac{2a}{2b}=\dfrac{3c}{3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c}{2b+3d}\)

\(\Rightarrow\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c}{2b+3d}\)

Equality occurs when and only when in three numbers a,b,c

+ There are 2 numbers equals 1

+ 1 number equals -1 

Good problem ! Thank you !

Use AM-GM for three numbers , we have :

a² + b² + c² \(\ge3.\sqrt[3]{a^2.b^2.c^2}\)

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

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

\(\Leftrightarrow1\ge a^2b^2c^2\)

\(\Leftrightarrow1\ge abc\)

\(\Leftrightarrow-abc\le1\)              (1) 

Now , we have to prove this .

(|a| + |b| + |c|)² \(\le\) 3(a² + b² + c²)

\(\Leftrightarrow\left|a\right|^2+\left|b\right|^2+\left|c\right|^2+2.\left|a\right|.\left|b\right|+2.\left|b.\right|.\left|c\right|+2.\left|c\right|.\left|a\right|\le3a^2+3b^2+3c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2.ab+2.bc+2.ca\le3a^2+3b^2+3c^2\)

\(\Leftrightarrow2.ab+2.bc+2.ca\le2a^2+2b^2+2c^2\)

\(\Leftrightarrow0\le2a^2+2b^2+2c^2-2ab-2bc-2ca\)

\(\Leftrightarrow0\le\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)           (right)

So , \(\left(\left|a\right|+\left|b\right|+\left|c\right|\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow\left(\left|a\right|+\left|b\right|+\left|c\right|\right)^2\le3.3=9\)

But |a| + |b| + |c| \(\ge\) 0

\(\Rightarrow\left|a\right|+\left|b\right|+\left|c\right|\le3\)                        (2)

From (1) and (2) 

=> |a| + |b| + |c| - abc \(\le\) 4

Use AM-GM for three numbers , we have :

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{\sqrt[3]{abc}}\)

So \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3.\sqrt[3]{abc}.\dfrac{3}{\sqrt[3]{abc}}=9\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

We have : 

\(\dfrac{2}{3}=\dfrac{2.1}{3.1}\)

\(\dfrac{4}{6}=\dfrac{2.2}{3.2}=\dfrac{2}{3}\)

\(\dfrac{6}{9}=\dfrac{2.3}{3.3}=\dfrac{2}{3}\)

..........

From the above analysis, we derive the general formula , is :

\(\dfrac{2}{3}=\dfrac{4}{6}=\dfrac{6}{9}=\dfrac{8}{12}=\dfrac{10}{15}=...........=\dfrac{2.k}{3.k}\) (with k > 0)

\(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{2.5}{2.\left(2x+1\right)}-\dfrac{3.2}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

<=> 2x + 1 = 13

<=> 2x = 12

<=> x = 6

We have : 

\(\dfrac{1}{51}>\dfrac{1}{100}\)

\(\dfrac{1}{52}>\dfrac{1}{100}\)

...............

\(\dfrac{1}{100}=\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+.....+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+....+\dfrac{1}{100}=\dfrac{50.1}{100}=\dfrac{1}{2}\)  (1)

In another case , we have :

\(\dfrac{1}{51}< \dfrac{1}{50}\)

\(\dfrac{1}{52}< \dfrac{1}{50}\)

.............

\(\dfrac{1}{100}< \dfrac{1}{50}\)

\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+.....+\dfrac{1}{100}< \dfrac{1}{50}+\dfrac{1}{50}+....+\dfrac{1}{50}=\dfrac{50.1}{50}=1\) (2)

From (1) and (2) 

=> \(\Rightarrow\dfrac{1}{2}< \dfrac{1}{51}+\dfrac{1}{52}+.....+\dfrac{1}{100}< 1\)

Equality occurs :

<=> x = y = z = 0 

This is a consequence of AM-GM (Cauchy) inequality 

Let ab = x , bc = y , ca = z

We have : \(\left(x+y+z\right)^2\ge3ab.ac+3bc.ab+3.ac.bc\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx\ge3.x.z+3.x.y+3.c.y\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2\ge0\) (right)

So , \(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)

Haizzzzzzzzzzz,,,,,,,,,,,,,,,,,,,,,,,,,, , whatever 

Let \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2006}}\)

\(2A=2+1+\dfrac{1}{2}+.....+\dfrac{1}{2^{2005}}\)

\(2A-A=2-\dfrac{1}{2^{2006}}\)

\(A=\dfrac{2^{2007}-1}{2^{2006}}\)

The last row will be :

\(A=\dfrac{2425}{4851}.\dfrac{1}{2}=\dfrac{2425}{9702}\)

Let \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{97.98.99}\)

\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.....+\dfrac{2}{97.98.99}\)

\(2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{97.98}-\dfrac{1}{98.99}\)

\(2A=\dfrac{1}{1.2}-\dfrac{1}{98.99}=\dfrac{1}{2}-\dfrac{1}{98.99}=\dfrac{9702}{2.98.99}-\dfrac{2}{2.98.99}=\dfrac{9700}{19404}=\dfrac{2425}{4851}\)

We have : triangle ABC and G is a centroid .

So AG , BG , CG cut BC , CA , AB at D , M , E

Draw AH \(\perp\) BM , CK \(\perp\) BM

We have : 

\(\Delta AHM=\Delta CKM\) (hypotenuse case - acute angle)

=> AH = CK

Two triangles GAB and GBC have GB is the common bottom 

AH = CK 

So , the area of \(S_{\Delta GAB}=S_{\Delta GAC}\)  (1)

Similar with  \(S_{\Delta GBC}=S_{\Delta GAC}\)   (2)

From (1) and (2) 

=> \(S_{\Delta GAB}=S_{\Delta GAC}=S_{\Delta GBC}\)

\(0\le\dfrac{a}{11}\le99\)

\(\Leftrightarrow0\le a\le1089\)

So , there are : (1089 - 0) : 1 + 1 = 1090 (numbers)

No further discussion , we will wait for the answer from the teacher .

If me or you're wrong, they both learn from experience . Ok ???

Phan Thanh Tinh : If we want to find the Maximum value of P

So , P must be positive 

Means x is a positive number 

=> x \(\ge0\)

Cauchy's inequality is also known as the AM-GM inequality