Đặt A=1/101+1/102+1/103+...+1/300

vì 1/101>1/102>1/103>...>1/300

=>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+(1/300+1/300+1/300+...+1/300) (mỗi ngoặc tên có tất cả là 100 phân số/1 ngoặc nhé!) 

=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100

=> A > 1/2+1/3

=> A > 5/6 

Mà 5/6>2/3

=> A > 2/3

Vậy 1/101+1/102+1/103+...+1/300 >2/3

Vì : 1/101 > 1/300 ;  1/102 > 1/300 .... ; 1/299 >1/300 ;    Do 1/101.....1/300 có 200 số 

=>1/101+1/102+....+1/299+1/300 > 1/300 x 200

                                                 >  2/3

1/101+1/102+...+1/299+1/300>2/3>1/300+1/300+1/300=200/300=2/3

vay 1/101+1/102+..+1/299+1/300>2/3

ta có 

\(\frac{1}{300}< \frac{1}{101}\); \(\frac{1}{300}< \frac{1}{102}\); \(\frac{1}{300}< \frac{1}{102}\)....\(\frac{1}{300}< \frac{1}{299}\)

\(\frac{1}{300}+\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}< \frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)

\(\frac{200}{300}< \frac{1}{101}+\frac{1}{102}+...+\text{​​}\text{​​}\)

rút gọn là xong

Đặt A=1/101+1/102+1/103+...+1/300

vì 1/101>1/102>1/103>...>1/300

=>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+(1/300+1/300+1/300+...+1/300) (mỗi ngoặc tên có tất cả là 100 phân số/1 ngoặc nhé!) 

=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100

=> A > 1/2+1/3

=> A > 5/6 

Mà 5/6>2/3

=> A > 2/3

Vậy 1/101+1/102+1/103+...+1/300 >2/3

Đặt A=1/101+1/102+1/103+...+1/300

vì 1/101>1/102>1/103>...>1/300

=>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+(1/300+1/300+1/300+...+1/300) (mỗi ngoặc tên có tất cả là 100 phân số/1 ngoặc nhé!) 

=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100

=> A > 1/2+1/3

=> A > 5/6 

Mà 5/6>2/3

=> A > 2/3

Vậy 1/101+1/102+1/103+...+1/300 >2/3

Đặt A=1/101+1/102+1/103+...+1/300

vì 1/101>1/102>1/103>...>1/300 =>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+ (1/300+1/300+1/300+...+1/300) (mỗi ngoặc tên có tất cả là 100 phân số/1 ngoặc nhé!)

=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100

=> A > 1/2+1/3

=> A > 5/6 Mà 5/6>2/3

=> A > 2/3 Vậy 1/101+1/102+1/103+...+1/300 >2/3

Ta có :

\(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}=VP\left(đpcm\right)\)

Xét :

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{200}\right)\)

Thêm \(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\)vào mỗi vế ta có

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)

\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)

\(\RightarrowĐPCM\)

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)( có 200 số )

Ta có

\(\frac{1}{101}>\frac{1}{300}\); \(\frac{1}{102}>\frac{1}{300}\); ...;\(\frac{1}{299}>\frac{1}{300}\)

=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}+\frac{1}{300}\)

=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{1}{300}.200\)

=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{2}{3}\)( dpcm )

Ta có\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>200.\frac{1}{300}=\frac{200}{300}=\frac{2}{3}\Rightarrowđpcm\)

Ta có: \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}+\frac{1}{300}>\frac{1}{300}.200=\frac{200}{300}=\frac{2}{3}\)

Vậy \(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{300}>\frac{2}{3}\)

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>200.\frac{1}{300}\)

                                                               \(>\frac{2}{3}\)

là sao ??

có tất cả 200 số hạng.

mà 1/300 x 200 = 2/3

có 1/101>1/300

    1/102>1/300

     ...

    1/299>1/300

    1/300=1/300

suy ra 1/101 + 1/102 + ... +1/299 +1/300 > 1/300 + 1/300 +...+ 1/300 + 1/300 =1/300 x 200=2/3

Ta có

\(\frac{1}{101}>\frac{2}{3}\)

\(\frac{1}{102}>\frac{2}{3}\)

.

.

.

\(\frac{1}{300}>\frac{2}{3}\)

Vậy \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)