Question

Cho a + b + c = 1 . Chứng minh rằng :

ab + bc + ca < \(\dfrac{1}{2}\)

Answer 1

Ta có:

\(a+b+c=1\)

\(\Rightarrow\left(a+b+c\right)^2=1\)

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=1\)

\(\Rightarrow2ab+2ac+2bc=1-a^2-b^2-c^2\)

\(\Rightarrow2\left(ab+ac+bc\right)=1-a^2-b^2-c^2\)

Vì \(1-a^2-b^2-c^2< 1\)

\(\Rightarrow2\left(ab+ac+bc\right)< 1\)

\(\Rightarrow ab+ac+bc< \dfrac{1}{2}\)

Answer 2

a + b + c =1 ⇔ (a + b + c)² = 1

⇔ a² + b² + c² + 2ab +2ac +2bc = 1

⇔2(ab + bc +ca) = 1 - a² + b² + c²

⇒2(ab + bc + ca) < 1

⇔ ab + bc +ca < \(\dfrac{1}{2}\)