\(B=\dfrac{1}{x^2+2}\le\dfrac{1}{2}\)
\("="\Leftrightarrow x=0\)
\(C=\dfrac{x^2+15}{x^2+3}=\dfrac{x^2+3+12}{x^2+3}=1+\dfrac{12}{x^2+3}\le1+\dfrac{12}{3}=5\)
\("="\Leftrightarrow x=0\)
\(D=\dfrac{x^2+y^2+5}{x^2+y^2+3}=\dfrac{x^2+y^2+3+2}{x^2+y^2+3}=1+\dfrac{2}{x^2+y^2+3}\le1+\dfrac{2}{3}=\dfrac{5}{3}\)
\("="\Leftrightarrow x=y=0\)