\(A=\dfrac{1}{z}\left(\dfrac{x+y}{xy}\right)=\dfrac{1}{z}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{4}{z\left(x+y\right)}\ge\dfrac{16}{\left(x+y+z\right)^2}=16\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{4};\dfrac{1}{4};\dfrac{1}{2}\right)\)

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

\(M=3\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)+\dfrac{1}{2xy}\ge\dfrac{12}{2xy+x^2+y^2}+\dfrac{2}{\left(x+y\right)^2}=\dfrac{14}{\left(x+y\right)^2}=14\)

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

Áp dụng bđt đã cho ta có \(M=4\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)-\dfrac{1}{x^2+y^2}\ge\dfrac{16}{2xy+x^2+y^2}-\dfrac{2}{\left(x+y\right)^2}=\dfrac{16}{\left(x+y\right)^2}-\dfrac{2}{\left(x+y\right)^2}=14\).

Đẳng thức xảy ra khi và chỉ khi \(x=y=\dfrac{1}{2}\)

\(A\ge\dfrac{\left(1+2\right)^2}{x+y}=9\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{3};\dfrac{2}{3}\right)\)

Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)

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 \(x=y=\dfrac{1}{2}\)

\(P=\dfrac{1}{2xy}+\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\ge\dfrac{1}{\dfrac{2.\left(x+y\right)^2}{4}}+\dfrac{4}{2xy+x^2+y^2}=\dfrac{6}{\left(x+y\right)^2}=6\)

\(P_{min}=6\) khi \(a=b=\dfrac{1}{2}\)

Cách khác:

Đặt $xy=t$. Bằng $AM-GM$ dễ thấy $t\leq \frac{1}{4}$

\(P=\frac{1}{xy}+\frac{1}{(x+y)^2-2xy}=\frac{1}{xy}+\frac{1}{1-2xy}=\frac{1}{t}+\frac{1}{1-2t}\)

\(=\frac{1}{t}-4+\frac{1}{1-2t}-2+6=\frac{(1-4t)(1-3t)}{t(1-2t)}+6\geq 6\) với mọi $t\leq \frac{1}{4}$

Vậy $P_{\min}=6$ khi $x=y=\frac{1}{2}$