\(=\dfrac{2019\cdot2020\left(2020\cdot10001-2019\cdot10001\right)}{2020\cdot2019\cdot2018\cdot10001}=\dfrac{10001\cdot1}{10001\cdot2018}=\dfrac{1}{2018}\)

Đề là gì vậy bạn !

Ta chứng minh 1 bổ đề sau: Với a;b lớn hơn hoặc bằng 1 thì \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

Thật vậy: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{a^2+b^2+2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(a^2+b^2+2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow a^2+a^3b+b^2+b^3a+2+2ab\ge2a^2+2b^2+2a^2b^2+2\)

\(\Leftrightarrow a^3b+b^3a+2ab-a^2-b^2-2a^2b^2\ge0\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)(đúng với a;b>=1)

Trở lại bđt trong bài: \(\frac{2019}{2019+x^2}+\frac{2019}{2019+y^2}\ge\frac{4038}{2019+xy}\)

\(\Leftrightarrow\frac{1}{2019+x^2}+\frac{1}{2019+y^2}\ge\frac{2}{2019+xy}\) bđt này tương tự với bđt vừa cm trong bài,với x;y là hoán vị của a;b và 2019 có vai trò như 1

cái gì đấy

đặt 2²⁰¹⁸ = a ; 3²⁰¹⁹ = b ; 5²⁰²⁰ = c

Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)

\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)

\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)

\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)

\(B< \frac{3}{4}\)

\(\Rightarrow A>1>\frac{3}{4}>B\)

Mình chỉ biết cách tính B thôi, đây nhé:

B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)

B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)

A>b

\(\frac{2019.2020-4040}{2017.2018+4034}\)=\(\frac{\left(2017+2\right).2020-4040}{2017.2018+2017.2}\)

=\(\frac{2017.2020+2.2020-4040}{2017.\left(2018+2\right)}\)

=\(\frac{2017.2020+4040-4040}{2017.2020}\)

=\(\frac{2017.2020+0}{2017.2020}\)

=\(\frac{1}{1}\)=1

Đặt \(2016=a\) biểu thức trên trở thành:

\(P=\dfrac{\left(a^2\left(a+10\right)+31\left(a+1\right)-1\right)\left(a\left(a+5\right)+4\right)}{\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)}=\dfrac{A}{B}\)

Với \(B=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)\)

Ta có: \(a^2\left(a+10\right)+31\left(a+1\right)-1=a^3+10a^2+31a+30\)

\(=a^3+5a^2+6a+5a^2+25a+30=a\left(a^2+5a+6\right)+5\left(a^2+5a+6\right)\)

\(=\left(a+5\right)\left(a^2+5a+6\right)=\left(a+5\right)\left(a^2+2a+3a+6\right)\)

\(=\left(a+5\right)\left(a+2\right)\left(a+3\right)\)

Và \(a\left(a+5\right)+4=a^2+5a+4=a^2+a+4a+4=\left(a+1\right)\left(a+4\right)\)

\(\Rightarrow A=\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)=B\)

\(\Rightarrow P=\dfrac{A}{B}=1\)