EZ, đề thanh hóa sáng nay ^^
Ta có: \(VT=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{abc}\)
\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
\(\Rightarrow VT\ge\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{7}{ab+bc+ca}\)
\(\ge\frac{9}{\left(a+b+c\right)^2}+\frac{7.3}{\left(a+b+c\right)^2}=30\)
Ta có:
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{3}{3abc}\)
\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{3abc}+\frac{2}{3abc}\)
\(=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{3abc}+\frac{2}{3abc}\)
\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{3ab}+\frac{1}{3ac}+\frac{1}{3bc}+\frac{2}{3abc}\)
\(\ge\frac{16}{a^2+b^2+c^2+3ab+3ac+3bc}+\frac{2}{3abc}\)
\(=\frac{16}{\left(a+b+c\right)^2+ab+ac+bc}+\frac{2}{3abc}\)
Mặt khác:\(ab+ac+bc\le\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)và: \(abc\le\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}\left(cosi\right)\)
Từ đó\(\Rightarrow VT\ge\frac{16}{1+\frac{1}{3}}+\frac{2}{3.\frac{1}{27}}=12+18=30\left(đpcm\right)\)
Dấu = xảy ra khi:
\(a=b=c=\frac{1}{3}\)
