Mình ko chắc lắm :

Áp dụng BĐT AM - GM ta có :

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

\(=\frac{x^2y^2+1}{y^2}.\frac{x^2y^2+1}{x^2}=\frac{x^4y^4+2x^2y^2+1}{x^2y^2}\)

\(=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)

\(\ge2\sqrt{x^2y^2.\frac{1}{256x^2y^2}}+\frac{255}{256.\left(xy\right)^2}+2\)

\(\ge2.\frac{1}{16}+\frac{255}{256.\left(\frac{\left(x+y\right)^2}{4}\right)^2}+2\)

\(=\frac{1}{8}+\frac{255}{256.\left(\frac{1}{4}\right)^2}+2=\frac{289}{16}\)

Khi \(x=y=\frac{1}{2}\)

Chúc bạn học tốt !!!

1. \(1=x^2+y^2\ge2xy\Rightarrow xy\le\frac{1}{2}\)

 \(A=-2+\frac{2}{1+xy}\ge-2+\frac{2}{1+\frac{1}{2}}=-\frac{2}{3}\)

max A = -2/3 khi x=y=\(\frac{\sqrt{2}}{2}\)

\(\frac{1}{xy}+\frac{1}{xz}=\frac{1}{x}\left(\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x}.\frac{4}{y+z}=\frac{4}{\left(4-t\right)t}=\frac{4}{4-\left(t-2\right)^2}\ge1\) với t = y+z => x =4 -t

\(A=x^2+y^2=\frac{\left(1^2+1^2\right)\left(x^2+y^2\right)}{2}\ge\frac{\left(1.x+1.y\right)^2}{2}=\frac{1}{2}\)A min = 1 khi x =y = 1/2

\(\sqrt{A}=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}=x+y=1\)( \(\sqrt{a+b}\le\sqrt{a}+\sqrt{b}\))

=> A\(\le1\) => Max A = 1 khi x =0;y =1 hoặc x =1 ; y =0

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

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

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

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

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

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

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

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

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

\(C_{min}=4\) khi \(x=y=2\)

Hoặc là:

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

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

\(M=x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+4\)

\(M=\left(1-2xy\right)+\dfrac{1-2xy}{\left(xy\right)^2}+4=\dfrac{1}{\left(xy\right)^2}-\dfrac{2}{xy}-2xy+5\\ \)đặt 1/xy= t \(\left(x+y\right)=1\Rightarrow xy\le\dfrac{1}{4}\Rightarrow t\ge4\)

\(M=t^2-2t-\dfrac{2}{t}+5\)

khi t > 1 hiển nhiên M luôn tăng khi t tăng => \(Mmin=M\left(4\right)=4.4-2.4-\dfrac{2}{4}+5=\dfrac{25}{2}\)

Đẳng thức khi t=4 => xy=1/4 => x=y=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 $x=y=\frac{1}{2}$

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 \(x=y=\frac{1}{2}\)