\(G=x^4+y^4-4(x^3+y^3)-20(x^2+y^2)-2x^2y^2+xy\\=(x^4-2x^2y^2+y^4)-4(x+y)^3+12xy(x+y)-20(x+y)^2+40xy+xy\\=(x^2-y^2)^2-4.5^3+12xy.5-20.5^2+41xy\\=(x-y)^2.(x+y)^2+101xy-1000\\=25(x-y)^2+101xy-1000\\=25x^2+51xy+25y^2-1000\\=25(x+y)^2+xy-1000\\=25.5^2+xy-1000\\=xy-375\)
Với mọi x, y ta có:
\((x-y)^2\ge0\\\Rightarrow x^2-2xy+y^2\ge0\\\Rightarrow (x+y)^2\ge4xy\\\Rightarrow xy\le \frac{(x+y)^2}{4}=\frac{5^2}{4}=\frac{25}{4}\\\Rightarrow xy-375\le \frac{25}{4}-375=-\frac{1475}{4}\)
\(\Rightarrow G\le-\frac{1475}{4}\)
Dấu "=" xảy ra khi: \(\begin{cases} x+y=5\\ x=y \end{cases} \Rightarrow x=y=\frac52\)
Vậy \(G_{max}=-\frac{1475}{4}\) tại \(x=y=\frac52\).
$\text{#}Toru$
