Question

Help me!!!!!!!

Answer 1

Sửa đề: >=14

Đặt t=xy+yz+xz

=>x^2+y^2+z^2=1-2t

\(xy+yz+xz< =\dfrac{x^2+y^2+y^2+z^2+z^2+x^2}{2}=x^2+y^2+z^2\)

=>\(t< =\dfrac{2\left(xy+yz+xz\right)+\left(x^2+y^2+z^2\right)}{3}=\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{1}{3}\)

3/t+2/1-2t>=14, 0<=t<=1/3

=>3-6t+2t>=14t(1-2t)

=>3-4t>=14t-28t^2

=>3(1-3t)^2+t^2>=0(đúng)

=>BĐT được chứng minh