Đặt \(\log_9x=\log_6y=\log_4\left(2x+y\right)=t\)

\(\Rightarrow\left\{{}\begin{matrix}x=9^t\left(1\right)\\y=6^t\left(2\right)\\2x+y=4^t\left(3\right)\end{matrix}\right.\)

Thay (1);(2) vào (3) ta được:

2.(3^t)²+2^t.3^t-(2^t)²=0

<=>2.3^t=2^t =>2.9^t=6^t=>2x=y=>x/y=1/2

\(x^3+y^3+3xy\left(x+y\right)+\dfrac{1}{27}-3xy\left(x+y\right)-xy=0\)

\(\Leftrightarrow\left(x+y\right)^3+\dfrac{1}{27}-3xy\left(x+y+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x+y+\dfrac{1}{3}\right)\left[\left(x+y\right)^2-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}\right]-3xy\left(x+y+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x^2+y^2-xy-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-\dfrac{1}{3}\right)^2+\left(y-\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow x=y=\dfrac{1}{3}\Rightarrow P=...\)

\(VT=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)

\(VT\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{25}{2}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

\(VT\ge3\sqrt[3]{\dfrac{x^3y^3z^3\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}}=3xyz\) (dpcm)

Ta có: 

\(\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{z}{xy}=\dfrac{1}{2}\left(\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{x}{yz}+\dfrac{z}{xy}+\dfrac{y}{zx}+\dfrac{z}{xy}\right)\ge\dfrac{1}{2}\left(\dfrac{2}{z}+\dfrac{2}{y}+\dfrac{2}{x}\right)\)

\(\Rightarrow P\ge\dfrac{1}{2}\left(x^2+y^2+z^2\right)+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)

\(\Rightarrow P\ge\dfrac{1}{2}\left(x^2+\dfrac{1}{x}+\dfrac{1}{x}\right)+\dfrac{1}{2}\left(y^2+\dfrac{1}{y}+\dfrac{1}{y}\right)+\dfrac{1}{2}\left(z^2+\dfrac{1}{z}+\dfrac{1}{z}\right)\)

\(\Rightarrow P\ge\dfrac{3}{2}\sqrt[3]{\dfrac{x^2}{x^2}}+\dfrac{3}{2}\sqrt[3]{\dfrac{y^2}{y^2}}+\dfrac{3}{2}\sqrt[3]{\dfrac{z^2}{z^2}}=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)

<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)

<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)

đặt x+y+z = t 

=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)

=> x=y=z 

ta lại có 

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

vì x=y=z  => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)