áp dụng bđt cosi cho 2 số dương, ta có:
\(x+y\ge2\sqrt{xy}\)
\(y+z\ge2\sqrt{yz}\)
\(x+z\ge2\sqrt{xz}\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{xz}=8\sqrt{x^2y^2z^2}=8xyz\)
Dấu "=" xảy ra khi \(x=y;y=z;x=z\Rightarrow x=y=z\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=8xyz\Leftrightarrow x=y=z\left(đpcm\right)\)
ta có:
\(x+y\ge2\sqrt{xy}\)
\(y+z\ge2\sqrt{yz}\)
\(x+z\ge2\sqrt{xz}\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge2\sqrt{yz}.2\sqrt{xz}\)
\(=8\sqrt{x^2y^2z^2}=8xyz\)
Ta thấy dấu = xảy ra khi:
\(x=y;y=z;x=z\Rightarrow x=y=z\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=8xyz\Rightarrow x=y=z\left(đpcm\right)\)
Nguyễn Thanh HằngAkai HarumaPhạm Hoàng GiangAn TrầnAn Trịnh HữuPhạm Tú UyênBùi Thị VânGửi tin nhắn