Có VT = \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{zx}}\)

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

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|=VP\) (Vì x + y + z = 0) 

Áp dụng BĐT BSC và BĐT Cosi:

\(17\left(x+y+z\right)+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\ge17\left(x+y+z\right)+\dfrac{2.\left(1+1+1\right)^2}{x+y+z}\)

\(=17\left(x+y+z\right)+\dfrac{18}{x+y+z}\)

\(=17\left(x+y+z\right)+\dfrac{17}{x+y+z}+\dfrac{1}{x+y+z}\)

\(\ge2\sqrt{17\left(x+y+z\right).\dfrac{17}{x+y+z}}+\dfrac{1}{1}\)

\(=35\)

\(\Rightarrow17\left(x+y+z\right)+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge35\)

Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}\)

\(17x+\dfrac{17}{9x}\ge\dfrac{34}{3}\)

tương tự.....

suy ra 

\(17\left(x+y+z\right)+\dfrac{17}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{34}{3}.3=34\)

lại có 

\(\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{9}{x+y+z}.\dfrac{1}{9}=1\)

nên

\(17\left(x+y+z\right)+\dfrac{17}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=17\left(x+y+z\right)+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge35\)

Ta có \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\)

\(\Leftrightarrow\dfrac{\left(yz\right)^2+\left(xz\right)^2+\left(xy\right)^2+2xyz}{\left(xyz\right)^2}=1\)

<=> (xy)² + (yz)² + (zx)² + 2xyz = (xyz)² 

<=> (xy)² + (yz)² + (xz)² + 2xyz(x + y + z) = (xyz)² 

<=> (xy + yz + zx)² = (xyz)² 

<=> \(\left[{}\begin{matrix}xy+yz+zx=xyz\\xy+yz+zx=-xyz\end{matrix}\right.\)

+) Khi xy + yz + zx = -xyz 

=> \(\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-1< 0\left(\text{loại}\right)\)

=> xy + yz + zx = xyz 

<=> \(xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=xyz\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1\right)=0\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)

<=> \(\dfrac{x+y}{xy}=\dfrac{-\left(x+y\right)}{\left(x+y+z\right)z}\)

<=> \(\left(x+y\right)\left(\dfrac{1}{xz+yz+z^2}+\dfrac{1}{xy}\right)=0\)

<=> \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(zx+yz+z^2\right)xy}=0\)

<=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)

Khi x = -y => y = 1 => P = 1

Tương tự y = -z ; z = -x được P = 1

Vậy P = 1 

Ta có:

\(VT=2+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{y}{z}+\dfrac{x}{z}+\dfrac{z}{x}\)

Do đó ta chỉ cần chứng minh:

\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)

Ta có:

\(\dfrac{x}{y}+\dfrac{x}{y}+1\ge3\sqrt[3]{\dfrac{x^2}{y^2}}\) 

Tương tự ...

Cộng lại ta có:

\(2\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\right)+6\ge3\left(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\right)\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\)

Do đó ta chỉ cần chứng minh:

\(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)

\(\Leftrightarrow\left(\sqrt[3]{\dfrac{x}{y}}-\sqrt[3]{\dfrac{x}{z}}\right)^2+\left(\sqrt[3]{\dfrac{y}{x}}-\sqrt[3]{\dfrac{y}{z}}\right)^2+\left(\sqrt[3]{\dfrac{z}{x}}-\sqrt[3]{\dfrac{z}{y}}\right)^2\ge0\) (luôn đúng)

Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:

\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).

Không mất tính tổng quát, giả sử x + y = 0

\(\Leftrightarrow x=-y\)

\(\Leftrightarrow x^3=-y^3\).

Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).

Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).

Lời giải:

Từ \(xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Đặt \((a,b,c)=\left(\frac{1}{x}; \frac{1}{y}; \frac{1}{z}\right)\Rightarrow a+b+c=1\)

BĐT cần chứng minh trở thành:

\(P=\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(c+1)(a+1)}\geq \frac{1}{16}(*)\)

Thật vậy, áp dụng BĐT Cauchy ta có:

\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)

\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq 3\sqrt[3]{\frac{a^3}{64^2}}=\frac{3a}{16}\)

\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq 3\sqrt[3]{\frac{b^3}{64^2}}=\frac{3b}{16}\)

Cộng theo vế các BĐT trên và rút gọn :

\(\Rightarrow P+\frac{a+b+c+3}{32}\geq \frac{3(a+b+c)}{16}\)

\(\Leftrightarrow P+\frac{4}{32}\geq \frac{3}{16}\Leftrightarrow P\geq \frac{1}{16}\)

Vậy \((*)\) được chứng minh. Bài toán hoàn tất.

Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)