a) Áp dụng BĐT Cauchy cho 2 số dương:
\(x^2+y^2\ge2\sqrt{\left(xy\right)^2}=2xy\)
\(y^2+z^2\ge2\sqrt{\left(yz\right)^2}=2yz\)
\(x^2+z^2\ge2\sqrt{\left(xz\right)^2}=2xz\)
Cộng từ vế của các BĐT trên:
\(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)
(Dấu "="\(\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\x=y\end{cases}}\Leftrightarrow x=y=z\))
b) \(2x^2+2y^2+z^2+2xy+2yz+2xz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+10x+25\right)\)
\(+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)(1)
Mà \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}}\)nên (1) xảy ra
\(\Leftrightarrow\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}z=8\\x=-5\\y=-3\end{cases}}\)
