Lời giải:

Từ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

$\Rightarrow xy+yz+xz=0$

Khi đó:

$x^2+2yz=x^2+yz-xz-xy=(x^2-xy)-(xz-yz)=x(x-y)-z(x-y)=(x-z)(x-y)$

Tương tự với $y^2+2zx, z^2+2xy$ thì:

$P=\frac{yz}{(x-z)(x-y)}+\frac{xz}{(y-z)(y-x)}+\frac{xy}{(z-x)(z-y)}$

$=\frac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}=\frac{-[yz(y-z)+xz(z-x)+xy(x-y)]}{-[xy(x-y)+yz(y-z)+xz(z-x)]}=1$

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

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

=>xy+yz+xz=0

=>yz=-xy-xz; xy=-xz-yz; xz=-xy-yz

\(x^2 + 2yz = x^2 + yz + yz = x^2 - x(y + z) + yz\)

\(=x^2-xy-xz+yz\)

\(=x(x-y)-z(x-y)=(x-y)(x-z)\)

Chứng minh tương tự, ta sẽ có:

\(y^2 + 2zx = (y - z)(y - x)\)

\(z^2 + 2xy = (z - x)(z - y)\)

Đặt \(A = \frac{1}{x^2 + 2yz} + \frac{1}{y^2 + 2zx} + \frac{1}{z^2 + 2xy}\)

\(=\frac{1}{(x - y)(x - z)}+\frac{1}{(y - z)(y - x)}+\frac{1}{(z - x)(z - y)}\)

\(=\frac{-1}{(x - y)(z - x)}+\frac{-1}{(y - z)(x - y)}+\frac{-1}{(z - x)(y - z)}\)

\(=\frac{-(y - z) - (z - x) - (x - y)}{(x - y)(y - z)(z - x)}\)

=0

=>\(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}=0\)

\(\left(\frac{1}{x^2 + 2yz}+\frac{1}{y^2 + 2zx}+\frac{1}{z^2 + 2xy}\right)\left(x^{2016}+y^{2017}+z^{2018}\right)\)

\(=0\left(x^{2016}+y^{2017}+z^{2018}\right)\)

=0

=xy+yz+xz

Áp dụng BĐT Cauchy-Schwarz, ta có:

\(VT\ge\dfrac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}=\dfrac{9}{\left(x+y+z\right)^2}=9\)

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

ĐK: \(x,y,z\ne0\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=0\Leftrightarrow xy+xz+yz=0\)

\(\Rightarrow\left\{{}\begin{matrix}xy=-xz-yz\\xz--xy-yz\\yz=-xy-xz\end{matrix}\right.\)

Ta có:

\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)\)

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

Tương tự: \(\dfrac{1}{y^2+2xz}=\dfrac{1}{\left(y-x\right)\left(y-z\right)}=\dfrac{-1}{\left(x-y\right)\left(y-z\right)}\)

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

Cộng vế với vế ta được:

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

\(=\dfrac{y-z-\left(x-z\right)+x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{y-z-x+z+x-y}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=0\) (đpcm)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow xy+yz+xz=0\)

\(A=\frac{yz}{x^2+yz+-xy-xz}+\frac{xz}{y^2+zx-xy-yz}+\frac{xy}{z^2+xy-xz-yz}\)

\(A=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-z\right)\left(y-x\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)

\(A=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)

\(A=\frac{\left(z-x\right)\left(y-z\right)\left(y-x\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}=1\)

@Unruly Kid

2 cái kìa còn lại làm tương tự rồi sau đó cộng lại với nhau sẽ ra 1 số tự nhiên nhé, dễ nên lười đánh nốt lắm :v

cam ơn ah. kết quả bằng 3 ah.