Question

Cho a,b,c>0 thoả mãn a+b+c=1. CMR:

\(\frac{1}{a^2+b^2+c^2}+\frac{1}{abc}\ge30\)

Answer 1

Ta có:

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

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

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

DO:

\(\frac{9}{\left(a+b+c\right)^2}+\frac{7}{ab+bc+ca}\ge9+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}=9+21=30\)

\(\Rightarrow DPCM\)

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