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Theo bđt Bunhiacopxki ta có : \(\left(x^2+y^2+z^2\right)\left(z^2+x^2+y^2\right)\ge\left(xy+yz+xz\right)^2\)
\(\Rightarrow x^2+y^2+z^2\ge\left|xy+yz+xz\right|\ge xy+yz+xz\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3\left(xy+yz+xz\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{3^2}{3}=3\) có GTLN là 3
Dấu "=" xảy ra khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
x2 + y2 - xy = 8
<=> 2x2 + 2y2 - 2xy = 16
<=> x2 + y2 + (x - y)2 = 16
<=> M = 16 - (x - y)2 \(\le\)16
Vậy max là 16
Ta lại có
2x2 + 2y2 - 2xy = 16
<=> 2x2 + 2y2 = 16 + 2xy
<=> 3(x2 + y2) = 16 + (x + y)2 \(\ge16\)
<=> 3M\(\ge\)16
<=> M \(\ge\frac{16}{3}\)
Vậy min là \(\frac{16}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y-x=1\Rightarrow x=y-1\)
\(\Rightarrow x^2+y^2=\left(y-1\right)^2+y^2\)
\(=y^2-2y+1+y^2\)
\(=2y^2-2y+1\)
\(=2\left(y^2-y+\frac{1}{2}\right)\)
\(=2\left(y^2-2y\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{2}\)
\(=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall y\)
Dấu"=" xảy ra khi \(2\left(y-\frac{1}{2}\right)^2=0\Rightarrow y=\frac{1}{2}\)
Vì \(y-x=1\)nên
\(\Rightarrow\frac{1}{2}-x=1\Rightarrow x=-\frac{1}{2}\)
Vậy \(Min_A=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2};y=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\le1\)