a, Xét : 3 - E = 3x^3-3xy-3y^3-x^3-xy-y^2/x^2-xy+y^2

= 2x^2-4xy+2y^2/x^2-xy+y^2

= 2.(x^2-2xy+y^2)/x^2-xy+y^2

= 2.(x-y)^2/x^2-xy+y^2 

>= 0 ( vì x^2-xy+y^2 > 0 )

Dấu "=" xảy ra <=> x-y=0 <=> x=y

Vậy ..........

b, Có : (x+1995)^2 = x^2+3990+1995^2 = (x^2-3990x+1995^2)+7980x

= (x-1995)^2 + 7980x >= 7980x

=> M < = x/7980x = 1/7980 ( vì x > 0 )

Dấu "=" xảy ra <=> x-1995=0 <=> x=1995

Vậy ...............

Ta có \(\frac{1}{P}=\frac{\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)^2}{x^3y^3}=\frac{x+yz}{y}\cdot\frac{y+zx}{x}\cdot\frac{\left(z+xy\right)^2}{x^2y^2}\)

\(=\left(\frac{x}{y}+z\right)\left(\frac{y}{x}+z\right)\left(\frac{z}{xy}+1\right)^2=\left[1+\left(\frac{x}{y}+\frac{x}{y}\right)z+x^2\right]\left(\frac{z}{xy}+1\right)^2\ge\left(1+2x+x^2\right)\)\(\left[\frac{4x}{\left(x+y\right)^2}+1\right]^2\)\(=\left(z+1\right)^2\left[\frac{4z}{\left(z-1\right)^2}+1\right]^2=\left[\frac{4z\left(z+1\right)}{\left(z-1\right)^2}+1\right]^2=\left[6+\frac{12}{z-1}+\frac{8}{\left(z-1\right)^2}+z-1\right]^2\)

\(=\left[6+\frac{12}{z-1}+\frac{3\left(z-1\right)}{4}+\frac{8}{\left(z-1\right)^2}+\frac{z-1}{8}+\frac{z-1}{8}\right]\)

Áp dụng BĐT Cosi ta có:

\(\frac{1}{P}\ge\left[6+2\sqrt{\frac{12}{z-1}\cdot\frac{3\left(z-1\right)}{3}}+3\sqrt[3]{\frac{8}{\left(z-1\right)^2}\cdot\frac{z-1}{8}\cdot\frac{z-1}{8}}\right]^2=\frac{729}{4}\)

\(\Rightarrow P\le\frac{4}{729}\). dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=2\\z=5\end{cases}}\)

Đặt \(\frac{\left(x+y+1\right)^2}{xy+x+y}=a\) ( ĐK a > 0 )

=> A = a + 1/a 

(*)  \(\left(x+y+1\right)^2\ge3\left(xy+x+y\right)\)( Nhân 2 vế với hai sau đưa về hằng đẳng thức ) 

=> \(\frac{\left(x+y+1\right)^2}{xy+x+y}\ge3\Leftrightarrow a\ge3\)

TA có \(A=a+\frac{1}{a}=\frac{a}{9}+\frac{1}{a}+\frac{8a}{9}\ge2\sqrt{\frac{a}{9}\cdot\frac{1}{a}}+\frac{8\cdot3}{9}=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)

Vậy GTNN của A là 10/3 tại x = y= 1 

\(x^3+3x^2+3x+1+y^3+3y^2+3y+1+x+y+2=0\)

\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0\)

\(\Leftrightarrow\left(x+y+2\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2+1\right]=0\)

\(\Leftrightarrow x+y+2=0\Rightarrow x+y=-2\)

Mà \(xy>0\Rightarrow\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\)

\(P=-\left(\frac{1}{-x}+\frac{1}{-y}\right)\le-\frac{4}{-x-y}=-2\)

\(P_{max}=-2\) khi \(x=y=-1\)

Đặt \(\frac{x}{y}+\frac{y}{x}=a\Rightarrow\frac{x^2}{y^2}+\frac{y^2}{x^2}=a^2-2\)

Ta cũng có: \(a=\frac{x^2+y^2}{xy}=\frac{\left(x-y\right)^2}{xy}+2\ge2\)

Vậy \(B=2\left(a^2-2\right)-a+1\) với \(a\ge2\)

\(B=2a^2-a-3=2a^2-a-6+3\)

\(B=\left(a-2\right)\left(2a+3\right)+3\)

Do \(a\ge2\Rightarrow\left\{{}\begin{matrix}a-2\ge0\\2a+3>0\end{matrix}\right.\) \(\Rightarrow\left(a-2\right)\left(2a+3\right)\ge0\)

\(\Rightarrow B\ge3\Rightarrow B_{min}=3\) khi \(a=2\) hay \(x=y\)