Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

 đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)

\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)

BBDT AM-GM 

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)

theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)

vì \(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(x^2+z^2\ge2xz\)

\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)

\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)

dấu"=" xảy ra<=>x=y=z=1/3

Ta có :

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

Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

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

\(=\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)

tương tự với hai ông còn lại sau đó cộng lại ta được:

\(\Sigma\dfrac{1}{3x+3y+2z}\le\dfrac{24}{16}=\dfrac{3}{2}\)

Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)

\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)

CMTT và cộng theo vế:

\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)

\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\Sigma\dfrac{1}{2x+3y+3z}\le\Sigma\dfrac{1}{16}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)

\(\Rightarrow P\le\dfrac{4}{16}\Sigma\left(\dfrac{1}{x+y}\right)=\dfrac{2017}{4}\)

Dấu " = " xảy ra khi \(x=y=z=\dfrac{3}{4034}\)

Áp dụng BĐT BSC:

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

\(\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}\right)\)

\(=\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)

\(maxF=1\Leftrightarrow x=y=z=\dfrac{3}{4}\)

\(TH_1:x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\\ \Rightarrow Q=\dfrac{-z}{z}+\dfrac{-x}{x}+\dfrac{-y}{y}=-3\\ TH_2:x+y+z\ne0\\ \Rightarrow\dfrac{3x-2y+z}{x}=\dfrac{3y-2z+x}{y}=\dfrac{3z-2x+y}{z}=\dfrac{2x+2y+2z}{x+y+z}=2\\ \Rightarrow\left\{{}\begin{matrix}3x-2y+z=x\\3y-2z+x=y\\3z-2x+y=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-2y=-z\\2y-2z=-x\\2z-2x=-y\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x-y=-\dfrac{z}{2}\\y-z=-\dfrac{x}{2}\\z-x=-\dfrac{y}{2}\end{matrix}\right.\)

\(\Rightarrow Q=-\dfrac{z}{2}:z-\dfrac{x}{2}:x-\dfrac{y}{2}:y=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)