Giả sử điều cần c/m là đúng , ta có :
\(\left(ab+bc+ac\right)^2\ge3abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2ab^2c+2a^2bc+2abc^2\ge3abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\ge3abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow2\left(a^2b^2+b^2c^2+a^2c^2\right)\ge2abc\left(a+b+c\right)\)
\(\Leftrightarrow2a^2b^2+2b^2c^2+2a^2c^2\ge2a^2bc+2ab^2c+2abc^2\)
\(\Leftrightarrow2a^2b^2+2b^2c^2+2a^2c^2-2a^2bc-2ab^2c-2abc^2\ge0\)
\(\Leftrightarrow\left(a^2b^2-2a^2bc+a^2c^2\right)+\left(b^2c^2-2ab^2c+a^2b^2\right)+\left(a^2c^2-2abc^2+b^2c^2\right)\ge0\)
\(\Leftrightarrow\left(ab-ac\right)^2+\left(bc-ab\right)^2+\left(ac-bc\right)^2\ge0\)
( điều này luôn đúng )
\(\Rightarrow\) điều giả sử là đúng
\(\Rightarrow\left(ab+bc+ac\right)^2\ge3abc\left(a+b+c\right)\)
