Question

x²+9y²+z²+\(\dfrac{19}{2}\) >2x+12y+4z

Answer 1

`x^2+9y^2+z^2+19/2 > 2x+12y+4z`

`<=>x^2-2x+1+9y^2-12y+4+z^2-4z+4+1/2 > 0`

`<=>(x-1)^2+(3y-2)^2+(z-2)^2 > -1/2` (LĐ `AA x,y,z`)

Answer 2

\(x^2+9y^2+z^2+\dfrac{19}{2}>2x+12y+4z\)

\(\left(x^2-2x+1\right)+\left(9y^2-12y+4\right)+\left(z^2-4z+4\right)+\dfrac{1}{2}>0\)

\(\left(x-1\right)^2+\left(3y-2\right)^2+\left(z-2\right)^2+\dfrac{1}{2}>0\) (luôn đúng)

Ta có: điều phải chứng minh