$1.$
[tex]Đặt \ a=2x+y+z;b=2y+z+x;c=2z+x+y\\\Rightarrow a+b+c=4x+4y+4z\Rightarrow x+y+z=\dfrac{a+b+c}{4}\\2x+y+z-(x+y+z)=x \ hay \ a-(x+y+z)=x\\\Rightarrow a-\dfrac{a+b+c}{4}=x\Leftrightarrow x=\dfrac{3a-b-c}{4}\\Tương \ tự \ ta \ có \ y=\dfrac{3b-c-a}{4};z=\dfrac{3c-a-b}{4}\\\Rightarrow D=\dfrac{3a-b-c}{4a}+\dfrac{3b-c-a}{4b}+\dfrac{3c-a-b}{4c}\\=\dfrac{3}{4}-\dfrac{b}{4a}-\dfrac{c}{4a}+\dfrac{3}{4}-\dfrac{c}{4b}-\dfrac{a}{4b}+\dfrac{3}{4}-\dfrac{a}{4c}-\dfrac{b}{4c}\\=\dfrac{9}{4}-\dfrac{1}{4}\left [\left ( \dfrac{a}{b}+\dfrac{b}{a} \right )+\left (\dfrac{b}{c}+\dfrac{c}{b} \right )+\left (\dfrac{a}{c}+\dfrac{c}{a} \right ) \right ]\\Áp \ dụng \ BĐT \ Cô-si \ cho \ 3 \ số \ dương \ a,b,c \ ta \ được:\\D\leq \dfrac{9}{4}-\dfrac{1}{4}(2+2+2)=\dfrac{3}{4}\Leftrightarrow a=b=c(đpcm)[/tex]