Ta có:

VT: \(\left(xy+1\right)\left(x^2y^2-xy+1\right)+\left(x^3-1\right)\left(1-y^3\right)\)

\(=\left(xy\right)^3+1^3+x^3-x^3y^3-1+y^3\)

\(=x^3y^3+1+x^3-x^3y^3-1+y^3\)

\(=\left(x^3y^3-x^3y^3\right)+\left(1-1\right)+\left(x^3+y^3\right)\)

\(=x^3+y^3=VP\left(dpcm\right)\)

a: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{yz+xz+xy}{xyz}=0\)

=>xy+xz+yz=0

Đặt a=xy; b=xz; c=yz

=>a+b+c=0

=>\(\left(a+b+c\right)^2=0\)

=>\(a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left\lbrack a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right\rbrack\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)

Ta có: \(\left(a^2+b^2+c^2\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

=>\(2\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2-\left(a^4+b^4+c^4\right)\)

=>\(4\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

DO đó, ta có: \(\left(a^2+b^2+c^2\right)^2=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

=>\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

=>\(2\left(x^4y^4+y^4z^4+x^4z^4\right)=\left(x^2y^2+z^2y^2+x^2z^2\right)^2\)

b:

x+y+z=0

=>x+y=-z

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)-3xzy+z^3\)

\(=\left(-z\right)^3-3xy\cdot\left(-z\right)-3xyz+z^3=z^3+3xyz-3xyz-z^3=0\)

\(A=3\cdot\dfrac{1}{8}\cdot\dfrac{-1}{3}+6\cdot\dfrac{1}{4}\cdot\dfrac{1}{9}+3\cdot\dfrac{1}{2}\cdot\dfrac{-1}{27}\)

\(=-\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{18}=\dfrac{-9}{72}+\dfrac{12}{72}-\dfrac{4}{72}=-\dfrac{1}{72}\)

Câu b đề sai rồi bạn

Lời giải:
a. Xét hiệu:

$x^3+y^3-xy(x+y)=(x^3-x^2y)-(xy^2-y^3)=x^2(x-y)-y^2(x-y)$

$=(x-y)(x^2-y^2)=(x-y)^2(x+y)\geq 0$ với mọi $x,y\geq 0$

$\Rightarrow x^3+y^3\geq xy(x+y)$

Dấu "=" xảy ra khi $x=y$

b.

Áp dụng BĐT phần a vô:

$x^3+y^3\geq xy(x+y)$

$\Rightarrow x^3+y^3+1\geq xy(x+y)+1=xy(x+y)+xyz=xy(x+y+z)$

$\Rightarrow \frac{1}{x^3+y^3+1}\leq \frac{1}{xy(x+y+z)}=\frac{xyz}{xy(x+y+z)}=\frac{z}{x+y+z}$

Hoàn toàn tương tự với các phân thức còn lại suy ra:

$\text{VT}\geq \frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}=1$

Ta có đpcm

Dấu "=" xảy ra khi $x=y=z=1$

Áp dụng bđt AM - GM:

\(x^3+1+1\ge3x;y^3+1+1\ge3y;z^3+1+1\ge3z;2x+2y+2z\ge6\sqrt[3]{xyz}=6\).

Cộng vế với vế các bđt trên rồi rút gọn ta có đpcm.

Áp dụng BĐT Cosi:

\(\left(x^3+1+1\right)+\left(y^3+1+1\right)+\left(z^3+1+1\right)\)

\(\ge3\left(x+y+z\right)\)

\(\ge x+y+z+2.3\sqrt[3]{xyz}\)

\(=x+y+z+6\)

\(\Rightarrow x^3+y^3+z^3\ge x+y+z\)

Đẳng thức xảy ra khi \(x=y=z=1\)

Ta có với x,y,z >0 thì:\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Bất đẳng thức Cô si ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x\sqrt{1-x^2}}\ge2\\ \Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\Leftrightarrow\dfrac{x^2}{\sqrt{1-x^2}}\ge2x^3\)
Tương tự: \(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Từ đó ta có:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)