I choose D because :
a<b => a-b<0 which means a-b is negative so I is true.

0<a<b => ab>0 => \(\dfrac{1}{ab}\) is positive so II is true.

\(\dfrac{1}{b}\) -\(\dfrac{1}{a}\)=\(\dfrac{a-b}{ab}\) but a-b<0 and ab>0 => \(\dfrac{a-b}{ab}\) is negative so III is false

Consider the expression :

\(S=\dfrac{1}{2^0}+\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{1992}{2^{1991}}\)

Prove that : S < 4

Compare two expressions A and B :

\(A=124.\left(\dfrac{1}{1.1985}+\dfrac{1}{2.1986}+\dfrac{1}{3.1987}+...+\dfrac{1}{16.2000}\right)\)

\(B=\dfrac{1}{1.17}+\dfrac{1}{2.18}+\dfrac{1}{3.19}+...+\dfrac{1}{1984.2000}\)

Prove that if a , b , c are positive numbers and a + b + c = 1 then :

\(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+\left(c+\dfrac{1}{c}\right)^2>33\)

Prove that if x + y = 1 and xy are different from 0 then :

\(\dfrac{y}{x^3-1}-\dfrac{x}{y^3-1}=\dfrac{2\left(x-y\right)}{x^2y^2+3}\)