Xài am-gm ta có:

\(\left(x-8y\right)+\frac{1}{y\left(x-8y\right)}+8y\ge3\sqrt[3]{\left(x-8y\right)\frac{1}{x-8y}8y}=6\)

Dấu = khi \(x-8y=\frac{1}{y\left(x-8y\right)}=8y\Leftrightarrow\hept{\begin{cases}x=4\\y=\frac{1}{4}\end{cases}}\)

\(\left(x-1;y-1\right)=\left(a;b\right)\Rightarrow\left\{{}\begin{matrix}a;b>0\\a+b\le2\end{matrix}\right.\)

\(A=\dfrac{\left(a+1\right)^4}{b^2}+\dfrac{\left(b+1\right)^4}{a^2}\ge\dfrac{1}{2}\left[\dfrac{\left(a+1\right)^2}{b}+\dfrac{\left(b+1\right)^2}{a}\right]^2\)

\(A\ge\dfrac{1}{2}\left[\dfrac{\left(a+b+2\right)^2}{a+b}\right]^2\ge\dfrac{1}{2}\left[\dfrac{8\left(a+b\right)}{a+b}\right]^2=32\)

\(T=\sqrt{\dfrac{x^3}{x^3+8y^3}}+\sqrt{\dfrac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(=\dfrac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\dfrac{2y^2}{\sqrt{y\left(y^3+\left(x+y\right)^3\right)}}\)

\(=\dfrac{x^2}{\sqrt{\left(x^2+2xy\right)\left(x^2-2xy+4y^2\right)}}+\dfrac{2y^2}{\sqrt{\left(xy+2y^2\right)\left(x^2+xy+y^2\right)}}\)

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

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

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

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

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

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

C₁:

\(x,y>0\)

\(M=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=x^2+2+\dfrac{1}{x^2}+y^2+2+\dfrac{1}{y^2}=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\)Theo BĐT AM-GM (Caushy) ta có:

\(M=\left(x^2+\dfrac{1}{16x^2}\right)+\left(y^2+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}.2\sqrt{\dfrac{1}{x^2}.\dfrac{1}{y^2}}+4=\dfrac{1}{2}+\dfrac{1}{2}+4+\dfrac{15}{4}.\dfrac{1}{xy}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{x+y}{2}\right)^2}\ge5+\dfrac{15}{4}.\dfrac{1}{\left(\dfrac{1}{2}\right)^2}=20\)Đẳng thức xảy ra \(\left\{{}\begin{matrix}x^2=\dfrac{1}{16}x^2\\y^2=\dfrac{1}{16}y^2\\x+y=1\\x,y>0\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy \(MinM=20\)