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

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

\(P=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{4xy}+4xy+\frac{5}{4xy}\)

\(P\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{\frac{4xy}{4xy}}+\frac{5}{\left(x+y\right)^2}=4+2+5=11\)

\(P_{min}=11\) khi \(x=y=\frac{1}{2}\)

Áp dụng BĐT Cauchy và Cauchy - Schwarz ta có:

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

\(=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(4xy+\frac{1}{4xy}\right)+\frac{5}{4xy}\)

\(\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{4xy\cdot\frac{1}{4xy}}+\frac{5}{\left(x+y\right)^2}\)

\(=\frac{4}{\left(x+y\right)^2}+2+\frac{5}{1^2}=4+2+5=11\)

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

1 thách dám tích

\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1}{2xy}\\ =\frac{1}{4}+\frac{1}{2xy}\ge\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)

Dấu = xảy ra khi x=y=2

\(\frac{1}{x^2}+\frac{1}{9y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{9y^2}}=\frac{2}{3xy}=\frac{2}{3}\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\frac{1}{x^2}=\frac{1}{9y^2}\\xy=1\end{cases}}\Rightarrow\hept{\begin{cases}x=\sqrt{3}\\y=\frac{1}{\sqrt{3}}\end{cases}}\).

\(P=\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}\)

\(\ge3\sqrt[3]{\frac{1}{xyz\left(x+1\right)\left(y+1\right)\left(z+1\right)}}\)

Mà theo BĐT AM - GM ta có tiếp:

\(xyz\le\left(\frac{x+y+z}{3}\right)^3=1\)

\(\left(x+1\right)\left(y+1\right)\left(z+1\right)\le\left(\frac{x+y+z+3}{3}\right)^3=8\)

\(\Rightarrow P\le\frac{3}{2}\)

Đẳng thức xảy ra tại x=y=z=1

Vậy..................