Ta cần chứng minh: \(\dfrac{a^2}{2}+b^2+c^2>ab+bc+ca\Leftrightarrow\dfrac{a^2}{2}+b^2+c^2-ab-bc-ca>0\Leftrightarrow\dfrac{a^2}{4}+b^2+c^2+ab+ca+2bc-3bc+\dfrac{a^2}{4}>0\) \(\Leftrightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2}{12}+\dfrac{a^2}{6}-3bc>0\Leftrightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2-36bc}{12}+\dfrac{a^2}{6}>0\) Mà \(a^3>36;abc=1\Rightarrow a^3>36abc\Rightarrow a^2>36bc\) 

\(\Rightarrow\left(\dfrac{a}{2}+b+c\right)^2+\dfrac{a^2-36bc}{12}+\dfrac{a^2}{6}>0\) luôn đúng

\(\Leftrightarrow x^2+\dfrac{5}{3}x-3=0\Leftrightarrow x^2+\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{133}{36}=0\Leftrightarrow\left(x-\dfrac{5}{6}\right)^2=\dfrac{133}{36}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{133}+5}{6}\\x=\dfrac{-\sqrt{133}+5}{6}\end{matrix}\right.\)

\(\Rightarrow D=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{2}{xy}\ge2\cdot\dfrac{4}{x^2+2xy+y^2}+\dfrac{2}{\dfrac{\left(x+y\right)^2}{4}}=\dfrac{4}{\left(x+y\right)^2}+\dfrac{8}{\left(x+y\right)^2}=\dfrac{4}{4}+\dfrac{8}{4}=3\) Dấu = xảy ra \(\Leftrightarrow x=y=1\)