Question

Cho a;b;c>0 thỏa mãn : a+b+c=1

Tìm min:

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\)

Answer 1

Theo C.B.S thì

\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\)

\(\Rightarrow\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ac}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)

Lại theo CBS thì

\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}=9\)mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)

\(\Rightarrow\dfrac{7}{ab+bc+ac}\ge21\)

\(\Rightarrow\)\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)\(\)\(\ge21+9=30\)

vậy Min = 30 khi a = b = c = 1/3