Question

Cho a,b,c là các số thực dương thỏa mãn a+b+c=1. Chứng minh:

\(\frac{1}{a^2+b^2+c^2}\)+\(\frac{1}{abc}\) \(\ge\) 30

Answer 1

\(VT=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

\(VT\ge\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)

\(VT\ge\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\)

\(VT\ge\frac{9}{\left(a+b+c\right)^2}+\frac{21}{\left(a+b+c\right)^2}=30\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)