Thanh Huynh

Nếu m = 0, ta có hệ :

$\left\{\begin{matrix} xy=0\\ 2y\sqrt[3]{x^{4}}=-\sqrt[3]{x^{4}} \end{matrix}\right.$

Hệ trên có nghiệm $\left ( x;y \right )=\left ( 0;c \right ), c\in R$ tùy ý

Xét m khác 0. Đặt $t=\sqrt[3]{x}$

$\left\{\begin{matrix} m\left ( t^{6}+t^{4}+t^{2}+1 \right )=yt^{3}\\ m\left ( t^{8}+t^{6}+t^{4}+t^{2}+1 \right )=\left ( 2y+1 \right )t^{2} \end{matrix}\right.$ $\left ( I \right )$

t=0 không là nghiệm của hệ

Đặt u = $t+\frac{1}{t}$ $\left ( |u|\geq 2 \right )$ hệ $\left ( I \right )$ trở thành :

$\left\{\begin{matrix} m\left ( u^{3}-2u \right )=y\\ m\left ( u^{4}-3u^{2}+1 \right )=2m\left ( u^{3}-2u \right )+1 \left ( * \right ) \end{matrix}\right.$

Hệ trên có nghiệm <=> pt (*) có nghiệm

hay $u^{4}-2u^{3}-3u^{2}+4u+1=\frac{1}{m}$

Đặt $f\left ( u \right )=u^{4}-2u^{3}-3u^{2}+4u+1$

Ta thấy với $|u|\geq 2$ thì $f\left ( u \right )\in [-3;+\infty ]$

Phương trình có nghiệm <=> $\frac{1}{m}\geq -3$

hay $m\geq 0$ hoặc m$\leq -\frac{1}{3}$

đặt x-1=m; y-1=n; z-1=p => mnp=1

P=$\frac{1}{\left ( m+1 \right )^{4}}+\frac{1}{\left ( n+1 \right )^{4}}+\frac{1}{\left ( p+1 \right )^{4}}$ $\geq \frac{1}{3}\left [ {\frac{1}{\left ( m+1 \right )^2}}+ {\frac{1}{\left ( n+1 \right )^2}}+{\frac{1}{\left ( p+1 \right )^2}}\right ]^{2}$

đặt $m=\frac{bc}{a^{2}}; n=\frac{ca}{b^{2}};p=\frac{ab}{c^{2}}$

ta có $\frac{1}{\left ( m+1 \right )^{2}}+\frac{1}{\left ( n+1 \right )^{2}}+\frac{1}{\left ( p+1 \right )^{2}}$

$=\frac{a^{4}}{\left ( a^{2} +bc\right )^{2}}+\frac{b^{4}}{\left ( b^{2} +ac\right )^{2}}+\frac{c^{4}}{\left ( c^{2} +ab\right )^{2}}$

$\geq \frac{a^{4}}{\left ( a^{2}+b^{2} \right )\left ( a^{2}+c^{2} \right )}+\frac{b^{4}}{\left ( b^{2}+c^{2} \right )\left ( a^{2}+b^{2} \right )}+\frac{c^{4}}{\left ( a^{2}+c^{2} \right )\left ( b^{2}+c^{2} \right )}$

$\geq \frac{\left ( a^{2}+b^{2}+c^{2} \right )^{2}}{\left (a^{2}+b^{2} \right )\left ( a^{2}+c^{2} \right )+\left ( b^{2}+c^{2} \right )\left ( a^{2}+b^{2} \right )+\left ( c^{2}+a^{2} \right )\left ( c^{2}+b^{2} \right )}$

$\geq \frac{3}{4}$

=> P$\geq \frac{1}{3}.\left ( \frac{3}{4} \right )^{2}=\frac{3}{16}$

dấu bằng xảy ra <=> x=y=z=2

Ta có : 

$\frac{x}{x^{2}+yz}\leq \frac{1}{4}\left ( \frac{1}{x}+\frac{x}{yz} \right )$

$\frac{y}{y^{2}+xz}\leq \frac{1}{4}\left ( \frac{1}{y}+\frac{y}{xz} \right )$

$\frac{z}{z^{2}+xy}\leq \frac{1}{4}\left ( \frac{1}{z}+\frac{z}{xy} \right )$

Vì $x^{2}+y^{2}+z^{2}\leq xyz$

<=>$\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\leq 1$

=>$\sum \frac{x}{x^{2}+yz}\leq \frac{1}{4}\left ( \frac{1}{x} +\frac{1}{y}+\frac{1}{z}\right )+\frac{1}{4}$

mà$\frac{1}{4}\left ( \frac{1}{x} +\frac{1}{y}+\frac{1}{z}\right )=\frac{1}{4}\frac{xy+yz+xz}{xyz}\leq \frac{1}{4}\frac{x^{2}+y^{2}+z^{2}}{xyz}\leq \frac{1}{4}$

$=> \sum \frac{x}{x^{2}+yz}\leq \frac{1}{2}$

dấu bằng xảy ra <=> x=y=z=3