Ta co:$8(x^{4}+y^{4})\geq4(x^{2}+y^{2})^{2}\geq(x+y)^{4}=1$(1) ( Theo $2(a^{2}+b^{2})\geq(a+b)^{2}$ )Mat khac: $1=(x+y)^{2}\geq4xy$$\Rightarrow$$\frac{1}{xy}\geq4$(2)
Cong (1) va (2) ve theo ve $\Rightarrow$ phuong trinh(2) co 1 nghiem duy nhat la x=y=$\frac{1}{2}$