Nếu \(a > b\) và $c < 0$ thì \(ac > bc\).

Nếu \(a < b\) và $c < 0$ thì \(ac > bc\).

Nếu \(a \ge b\) và $c < 0$ thì \(ac \le bc\).

Nếu \(a \ge b\) và $c > 0$ thì \(ac \ge bc\).

\(x < y\)                   

\(x = y\)            

\(x > y\)          

\(x \le y\)

\(a - b > 0\)                   

\(a - b < 0\)            

\(a - b = 0\)      

\(a - b \le 0\)

\(1\)    

\(2\)    

\(3\)    

\(0\)

\(2a + 2 > 2b + 4\)

\(2a + 2 < 2b + 4\)     

\(2a + 2 \ge 2b + 4\)

\(2a + 2 \le 2b + 4\)

\(\dfrac{{{a^2} + {b^2}}}{2} < ab\)

\(\dfrac{{{a^2} + {b^2}}}{2} \le ab\) 

\(\dfrac{{{a^2} + {b^2}}}{2} \ge ab\)

\(\dfrac{{{a^2} + {b^2}}}{2} > ab\)

\({\left( {x + y} \right)^2} \le 4xy\)                                               

\({\left( {x + y} \right)^2} > 4xy\)                                                

 \({\left( {x + y} \right)^2} < 4xy\)

\({\left( {x + y} \right)^2} \ge 4xy\)

\({a^2} + 5 > 4a\)                                             

\({a^2} + 10 < 6a - 1\)                                    

\({a^2} + 1 > a\)

 \(ab - {b^2} \le {a^2}\)

\({a^2} + {b^2} + {c^2} < ab + bc + ca\)    

\({a^2} + {b^2} + {c^2} \ge ab + bc + ca\)

\({a^2} + {b^2} + {c^2} \le ab + bc + ca\)

Cả A, B, C đều sai

\({a^3} + {b^3} - a{b^2} - {a^2}b < 0\)

\({a^3} + {b^3} - a{b^2} - {a^2}b \ge 0\)    

\({a^3} + {b^3} - a{b^2} - {a^2}b \le 0\)     

\({a^3} + {b^3} - a{b^2} - {a^2}b > 0\)