\(A=\dfrac{1}{a^2+b^2+c^2}+\dfrac{2018}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{4036}{2\left(ab+bc+ca\right)}\)Áp dụng BĐT cauchy schwarz ta có:
\(A=\dfrac{1}{a^2+b^2+c^2}+\dfrac{\left(\sqrt{4036}\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{\left(1+\sqrt{4036}\right)^2}{\left(a+b+c\right)^2}\ge\left(\dfrac{1+\sqrt{4036}}{3}\right)^2\)
Đẳng thức xảy ra khi và chỉ khi a=b=c=1