Lời giải:
Áp dụng BĐT Cô-si:

$x^3+1+1\geq 3x$

$y^3+1+1\geq 3y$

$z^3+1+1\geq 3z$

$\Rightarrow x^3+y^3+z^3+6\geq 3(x+y+z)\geq 3.3=9$

$\Rightarrow A=x^3+y^3+z^3\geq 3$ 

Vậy $A_{\min}=3$. Giá trị này đạt tại $x=y=z=1$

\(A=\left(x^3+1+1\right)+\left(y^3+1+1\right)+\left(z^3+1+1\right)-6\)

\(A\ge3\sqrt[3]{x^3}+3\sqrt[3]{y^3}+3\sqrt[3]{z^3}-6=3\left(x+y+z\right)-6\ge3.3-6=3\)

\(A_{min}=3\) khi \(x=y=z=1\)

Do \(x^2+y^2+z^2=1\Rightarrow x^2< 1\Rightarrow x< 1\)

\(\Rightarrow x^5< x^2\)

Tương tự ta có: \(y< 1\Rightarrow y^6< y^2\); \(z< 1\Rightarrow z^7< z^2\)

\(\Rightarrow x^5+y^6+z^7< x^2+y^2+z^2\)

\(\Rightarrow x^5+y^6+z^7< 1\)

Với mọi x;y;z ta luôn có:

\(\left(x+y-1\right)^2+\left(z-\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2+2xy-2x-2y+1+z^2-z+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow x^2+y^2+z^2+\dfrac{5}{4}+2xy-2x-2y-z\ge0\)

\(\Leftrightarrow2+2xy-2x-2y\ge z\)

\(\Leftrightarrow2\left(1-x\right)\left(1-y\right)\ge z\)

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

Ta có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow VT\le\dfrac{1}{xy\left(x+y\right)+xyz}+\dfrac{1}{yz\left(y+z\right)+xyz}+\dfrac{1}{zx\left(z+x\right)+xyz}\)

\(\Rightarrow VT\le\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=\dfrac{1}{x+y+z}.\left(\dfrac{x+y+z}{xyz}\right)=\dfrac{1}{xyz}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\)