Ta thấy 1/101>1/150 ; 1/102>1/150 ; .... ; 1/149>1/150 ; 1/150=1/150

suy ra 1/101+1/102+1/103+.....+1/149+1/150>50.1/150

1/101+1/102+1/103+.....+1/148+1/150>1/3

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

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

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

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

\(\RightarrowĐPCM\)

Tách A thành 2 nhóm A₁ , A₂

A₁ = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\)

A₂ = \(\frac{1}{151}+\frac{1}{152}+\frac{1}{153}+...+\frac{1}{200}>\frac{1}{200}.50=\frac{1}{4}\)

\(\Rightarrow\)A = A₁ + A₂ > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

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 : 

\(\frac{1}{101}>\frac{1}{200}\)

\(\frac{1}{102}>\frac{1}{200}\)

\(\frac{1}{103}>\frac{1}{200}\)

\(..........\)

\(\frac{1}{200}=\frac{1}{200}\)

Cộng vế với vế ta được :

\(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\) (có 100 số \(\frac{1}{200}\) )\(=\frac{100}{200}=\frac{1}{2}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+......+\frac{1}{200}>\frac{1}{2}\) (đpcm)

Ta có:

1/101>1/200

1/102>1/200

...

1/199>1/200

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

                                     =1/200.100=1/2

Vậy 1/101+1/102+1/103+...+1/200>1/2

\(\frac{1}{101}>\frac{1}{200}\)

\(\frac{1}{101}>\frac{1}{200}\)

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

\(\frac{1}{200}=\frac{1}{200}\)

Cộng vế với vế ta đc:

\(\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{200}>\frac{1}{200}+...+\frac{1}{200}\)(có 100 phân số \(\frac{1}{200}\))=\(\frac{1}{200}=\frac{1}{2}\)

=>...................