Lời giải:

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

$\Rightarrow (\frac{1}{x}+\frac{1}{y})+(\frac{1}{z}-\frac{1}{x+y+z})=0$

$\Leftrightarrow \frac{x+y}{xy}+\frac{x+y}{z(x+y+z)}=0$

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

$\Leftrightarrow (x+y).\frac{z(x+y+z)+xy}{xyz(x+y+z)}=0$

$\Leftrightarrow (x+y).\frac{(z+x)(z+y)}{xyz(x+y+z)}=0$

$\Leftrightarrow (x+y)(y+z)(x+z)=0$

$\Leftrightarrow x=-y$ hoặc $y=-z$ hoặc $z=-x$

Nếu $x=-y$ thì:

$P=\frac{3}{4}+[(-y)^8-y^8](y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}+0.(y^9+z^9)(z^{10}-x^{10})=\frac{3}{4}$

Nếu $y=-z$ thì:

$P=\frac{3}{4}+(x^8-y^8)[(-z)^9+z^9](z^{10}-x^{10})=\frac{3}{4}+(x^8-y^8).0.(z^{10}-x^{10})=\frac{3}{4}$

Nếu $z=-x$ thì:

$P=\frac{3}{4}+(x^8-y^8)(y^9+z^9)[(-x)^{10}-x^{10}]=\frac{3}{4}+(x^8-y^8)(y^9+z^9).0=\frac{3}{4}$

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

Nếu x+y=0 => x=-y

Nếu

\(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}=0\\ \Rightarrow xz+yz+z^2+xy=0\\ \Rightarrow\left(x+z\right)\left(y+z\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-z\\y=-z\end{matrix}\right.\)

Tự thế vào :v

Ta có \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\left(xy+xz+yz\right)\left(x+y+z\right)=xyz\Rightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\Rightarrow\left[{}\begin{matrix}x+y=0\\x+z=0\\y+z=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-y\\z=-x\\y=-z\end{matrix}\right.\)TH1: Nếu x=-y⇒x⁸-y⁸=x⁸-(-x)⁸=0 (Vì x⁸ và (-x)⁸ đều là số nguyên dương)⇒M=\(\text{​​}\dfrac{3}{4}+\left(x^8-y^8\right)\left(y^9-z^9\right)\left(z^{10}-x^{10}\right)=\dfrac{3}{4}\)

Tương tự với y=-z và z=-x

Vậy M=\(\dfrac{3}{4}\)

Từ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)

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

\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)

\(\Rightarrow\left(x+y\right)\left(\dfrac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)

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

Ta có: x⁸ - y⁸ = (x + y)(x - y)(x² + y²)(x⁴ + y⁴)

y⁹ + z⁹ = (y + z)(y⁸ - y⁷z + y⁶z² - ... + z⁸)

z¹⁰ - x¹⁰ = (z + x)(z⁴ - z³x + z²x² - zx³ + z⁴)(z⁵ - x⁵)

Vậy M = \(\dfrac{3}{4}\) + (x + y)(y + z)(z + x) = \(\dfrac{3}{4}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

(y + z - x)/x = (z + x - y)/y = (x + y - z)/z = 1

--> y + z - x = x; z + x - y = y; x + y - z = z

--> y + z = 2x; z + x = 2y; x + y = 2z

Ta có: 

B = (x + y)/y.(y + z)/z.(z + x)/x

= 2z/y.2x/z.2y/x = 8

Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)

\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)

\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)

từ đề bài ta có bất đẳng thức cần chứng minh tương đương: 

\(3+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{9}{4}\)

<=>\(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

ta có \(\dfrac{3}{4}+\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z}\le\dfrac{3}{4}+\dfrac{z+y}{4x}+\dfrac{x+z}{4y}+\dfrac{x+y}{4z}=\dfrac{3}{4}+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{4}=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\left(đpcm\right)\)Dấu "=" xảy ra khi x=y=z=\(\dfrac{1}{3}\)