Question

Cho a,b,c >0 thoả mãn: a+b+c=1. CMR: A= \(\dfrac{1}{abc}\)+\(\dfrac{1}{a^2 + b^2 + c^2}\)≥30

Answer 1

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

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

\(A\ge\dfrac{9}{ab+bc+ca+ab+bc+ca+a^2+b^2+c^2}+\dfrac{7}{\dfrac{1}{3}\left(a+b+c\right)^2}\)

\(A\ge\dfrac{30}{\left(a+b+c\right)^2}=30\)

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