Question

Chứng minh rằng với mọi a,b,c thì : 

\(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)

 

Answer 1

Đặt \(x=a+b+c;y=ab+bc+ac;z=abc\)

Suy ra : \(2\left(1+abc\right)+\sqrt{2\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)

\(\Leftrightarrow2\left(1+z\right)+\sqrt{2\left(x^2+y^2+z^2-2xz-2y+1\right)}\ge x+y+z+1\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-2xz-2y+1\right)\ge\left(x+y-z-1\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2-2xy-2xz+2x+2yz-2y-2z+1\ge0\)

\(\Leftrightarrow\left(x-y-z+1\right)^2\ge0\) (luôn đúng)

Vậy bđt ban đầu được chứng minh