Áp dụng BĐT AM-GM ta có:

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+\frac{1}{x}+y+\frac{1}{y}\right)^2}{2}\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{\left(1+4\right)^2}{2}=\frac{5^2}{2}=\frac{25}{2}\)

Xảy ra khi \(x=y=\frac{1}{2}\)

áp dùng BDT cô si chúa Pain có

\(\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\Rightarrow xy\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\ge2.\)

mà \(\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{2}\)

\(\Rightarrow\frac{xy}{2}\ge\Rightarrow xy\ge4\)

b)

áp dụng BDT cô si ta có

\(x+y\ge2\sqrt{xy}\)

lấy từ câu A ta có \(xy\ge4\) " câu a"

suy ra

\(x+y\ge2\sqrt{4}=4\)

\(\left(x-y\right)^2\ge0;\forall xy\Rightarrow x^2+y^2\ge2xy\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow x+y\ge2\sqrt{xy}\)

\(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{2}{xy}\Rightarrow xy\ge4\Rightarrow x+y\ge2\sqrt{xy}\ge2\sqrt{4}=4\)

\(C_{min}=4\) khi \(x=y=2\)

Hoặc là:

\(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(\dfrac{4}{x+y}\right)^2=\dfrac{8}{\left(x+y\right)^2}\)

\(\Rightarrow\left(x+y\right)^2\ge16\Rightarrow x+y\ge4\)

Ta có \(B\ge\dfrac{\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2}{2}\) \(=\dfrac{\left(1+\dfrac{1}{xy}\right)^2}{2}\)

Lại có \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\)

\(\Rightarrow B\ge\dfrac{\left(1+4\right)^2}{2}=\dfrac{25}{2}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Vậy GTNN của B là \(\dfrac{25}{2}\) khi \(x=y=\dfrac{1}{2}\)

Đề bài sai, C không có giá trị nhỏ nhất

Nếu \(\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{2}\) thì có thể tìm được min của C

\(1=2\left(\dfrac{9}{x}+\dfrac{1}{y}\right)\ge2.\dfrac{\left(3+1\right)^2}{x+y}=\dfrac{32}{x+y}\)

\(\Rightarrow x+y\ge32\)

\(A_{min}=32\) khi \(\left(x;y\right)=\left(24;8\right)\)

\(P=\left(2x+\dfrac{1}{x}\right)^2+9+\left(2y+\dfrac{1}{y}\right)^2+9-18\)

\(P\ge2\sqrt{9\left(2x+\dfrac{1}{x}\right)^2}+2\sqrt{9\left(2y+\dfrac{1}{y}\right)^2}-18\)

\(P\ge12x+12y+\dfrac{6}{x}+\dfrac{6}{y}-18\)

\(P\ge6\left(4x+\dfrac{1}{x}\right)+6\left(4y+\dfrac{1}{y}\right)-12\left(x+y\right)-18\)

\(P\ge6.2\sqrt{\dfrac{4x}{x}}+6.2\sqrt{\dfrac{4y}{y}}-12.1-18=18\)

\(P_{min}=18\) khi \(x=y=\dfrac{1}{2}\)