Question

 Cho a,b,c thoa

a^2+b^2+c^2=3

cm: ab+bc+ca+a+b+c<=6

 

Answer 1

Ta có : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\)

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

Ta cũng có : \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2-2\left(a+b+c\right)+3\ge0\)

\(\Leftrightarrow a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)

\(\Rightarrow\frac{a^2+b^2+c^2+3}{2}\ge a+b+c\)(2)

Cộng vế với vế ta được :

\(ab+bc+ac+a+b+c\le a^2+b^2+c^2+\frac{a^2+b^2+c^2+3}{2}=3+\frac{3+3}{2}=6\)