Ta có: \(\left(1\right)\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\). Dễ dàng suy từ BĐT Cô-si.
a, Ta có:
\(P\ge\frac{9}{a^2+2bc+b^2+2ca+c^2+2ab}=\frac{9}{\left(a+b+c\right)^2}\ge9\)
b, \(VT\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
\(=\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\right)+\frac{7}{ab+bc+ca}\)
\(\ge\frac{9}{\left(a+b+c\right)^2}+\frac{7}{ab+bc+ca}\ge\frac{9}{1}+\frac{7}{\frac{1}{3}}=30\)
Chú ý: \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)
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