Question

Cho x,y,z>0 và x+y+z\(\le\)3.

Chứng minh: \(\frac{1}{a^2+b^2+c^2}+\frac{2018}{ab+bc+ca}\ge673\)

Answer 1

Sao đang x, y, z lại sang a, b, c vậy

Answer 2

\(\frac{1}{a^2+b^2+c^2}+\frac{2018}{ab+bc+ca}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+ac+bc}+\frac{2017}{ab+ac+bc}\)

\(=\frac{1}{a^2+b^2+c^2}+\frac{2}{2ab+2ac+2bc}+\frac{2017}{ab+ac+bc}\ge\frac{\left(1+2\right)^2}{\left(a+b+c\right)^2}+\frac{2017}{\frac{\left(a+b+c\right)^2}{2}}=\frac{9}{3}+\frac{2017.2}{9}\)