Toán 10 Tìm GTNN

a) [tex]A=\frac{x^2+y^2}{xy}+\frac{xy}{x^2+y^2}[/tex]
Đặt [tex]t=\frac{x^2+y^2}{xy}\geq 2\Rightarrow A=t+\frac{1}{t}=\frac{t}{4}+\frac{1}{t}+\frac{3t}{4}\geq 2\sqrt{\frac{t}{4}.\frac{1}{t}}+\frac{3.2}{4}=\frac{5}{2}[/tex]
Dấu "=" xảy ra khi [TEX]x=y[/TEX]
b) [tex]B=\frac{x^2+y^2}{xy}+\frac{xy}{x^2+xy+y^2}=\frac{x^2+xy+y^2}{xy}+\frac{xy}{x^2+xy+y^2}-1[/tex]
Đặt [tex]t=\frac{x^2+xy+y^2}{xy}\geq 3[/tex]
[tex]B=t+\frac{1}{t}-1=\frac{t}{9}+\frac{1}{t}+\frac{8t}{9}-1\geq 2\sqrt{\frac{t}{9}.\frac{1}{t}}+\frac{8.3}{9}-1=\frac{7}{3}[/tex]
Dấu "=" xảy ra khi [TEX]x=y[/TEX]
c) [tex]C=\frac{(x+y)^2}{xy}+\frac{6xy}{(x+y)^2}-4[/tex]
Đặt [tex]t=\frac{(x+y)^2}{xy}\geq 4\Rightarrow C=\frac{6}{t}+t-4=(\frac{3t}{8}+\frac{6}{t})+\frac{5t}{8}-4\geq 2\sqrt{\frac{3t}{8}.\frac{6}{t}}+\frac{5.4}{8}-4=\frac{3}{2}[/tex]
Dấu "=" xảy ra khi [TEX]x=y[/TEX]
d) Ta có: [tex](x+y+1)^2=3(xy+x+y)+(x-1)^2+(y-1)^2+(x-y)^2\geq 3(xy+x+y)[/tex]
Đặt [tex]t=\frac{(x+y+1)^2}{xy+x+y}\geq 3\Rightarrow D=t+\frac{1}{t}=\frac{t}{9}+\frac{1}{t}+\frac{8t}{9}\geq 2\sqrt{\frac{t}{9}.\frac{1}{t}}+\frac{8.3}{9}=\frac{10}{3}[/tex]
Dấu "=" xảy ra khi [TEX]x=y=1[/TEX]