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\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

\(VT\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}\)

\(VT\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\le\dfrac{1}{2}\sqrt{3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)}=\dfrac{\sqrt{3}}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)

Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)

  \(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)

Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)

                         \(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

Cộng vế với vế ta có:

\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)

\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)

Áp dụng bất đẳng thức AM - GM:

\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).

Áp dụng bất đẳng thức AM - GM ta có:

\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).

Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).

Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)

\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).

Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)

Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)

\(\Rightarrow P\ge\dfrac{15}{2}\).

Vậy...

Áp dụng bất đẳng thức AM - GM:

P≥33√(xy+1)(yz+1)(zx+1)xyz.

Áp dụng bất đẳng thức AM - GM ta có:

xy+1=xy+14+14+14+14≥55√xy44.

Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.

Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412

⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.

Mà xyz≤(x+y+z)327=18

Nên  (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258 

⇒P≥152.

Đặt \(\left(\sqrt{x};2\sqrt{y};3\sqrt{z}\right)=\left(a;b;c\right)\Rightarrow a;b;c\ge0\)

Ta có:

\(\dfrac{2}{a+b+c}-\dfrac{1}{ab+bc+ca}\le\dfrac{2}{a+b+c}-\dfrac{3}{\left(a+b+c\right)^2}=-3\left(\dfrac{1}{a+b+c}-\dfrac{1}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)

Đẳng thức xảy ra khi và chỉ khi: \(a=b=c=1\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{4}\\z=\dfrac{1}{9}\end{matrix}\right.\)

Với a,b,c dưog thì \(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}>=\dfrac{\left(x+y+z\right)^2}{a+b+c}\)

\(P>=\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+\sqrt{1+x^3}+\sqrt{1+y^3}+\sqrt{1+z^3}}\)

\(\sqrt{1+x^3}=\sqrt{\left(1+x\right)\left(1-x+x^2\right)}< =\dfrac{2+x^2}{2}\)

Dấu = xảy ra khi x=2

=>\(P>=\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2+6}=\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2+6}\)

Đặt t=(x+y+z)^2(t>=36)

=>P>=2t/t-6

Xét hàm số \(f\left(t\right)=\dfrac{t}{t+6}\left(t>=36\right)\)

\(f'\left(t\right)=\dfrac{6}{\left(t+6\right)^2}>=0,\forall t>=36\)

=>f(t) đồng biến

=>f(t)>=f(36)=6/7

=>P>=12/7

Dấu = xảy ra khi x=y=z=2