\(BDT\Leftrightarrow\left(ab+bc+ac\right)^2\ge3a^2bc+3ab^2c+3abc^2\)
Đặt \(x=ab;y=bc;z=ac\) thì có:
\(\left(x+y+z\right)^2\ge3xy+3yz+3xz\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (đúng)
Đẳng thức xảy ra khi \(a=b=c\)
\(\left(ab+ac+bc\right)^2\ge3abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)-3abc\left(a+b+c\right)\ge0\)\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2-a^2bc-ab^2c-abc^2\ge0\)Nhân cả 2 vế cho 2 ta được
\(\Rightarrow2a^2b^2+2a^2c^2+2b^2c^2-2a^2bc-2ab^2c-2abc^2\ge0\)\(\Leftrightarrow\left(a^2b^2-2a^2bc+a^2c^2\right)+\left(a^2b^2-2ab^2c+b^2c^2\right)+\left(a^2c^2-2abc^2+b^2c^2\right)\ge0\)\(\Rightarrow\left(ab-ac\right)^2+\left(ab-bc\right)^2+\left(ac-bc\right)^2\ge0\) Đúng với mọi a , b , c
