Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{a^2}{b+2}+\frac{b+2}{9}\geq 2\sqrt{\frac{a^2}{9}}=\frac{2a}{3}\)
\(\frac{b^2}{c+2}+\frac{c+2}{9}\geq 2\sqrt{\frac{b^2}{9}}=\frac{2b}{3}\)
\(\frac{c^2}{a+2}+\frac{a+2}{9}\geq 2\sqrt{\frac{c^2}{9}}=\frac{2c}{3}\)
Cộng theo vế các BĐT thu được:
\(P+\frac{a+b+c+6}{9}\geq \frac{2}{3}(a+b+c)\)
\(\Leftrightarrow P\geq \frac{5}{9}(a+b+c)-\frac{6}{9}(*)\)
Áp dụng BĐT AM-GM:
\(3abc=a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow abc\geq \sqrt[3]{a^2b^2c^2}\Rightarrow a^3b^3c^3\geq a^2b^2c^2\)
\(\Rightarrow a^2b^2c^2(abc-1)\geq 0\Rightarrow abc\geq 1\)
Áp dụng BĐT AM-GM:
\(a+b+c\geq 3\sqrt[3]{abc}\geq 3\sqrt[3]{1}=3(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{5}{9}(a+b+c)-\frac{6}{9}\geq \frac{5}{9}.3-\frac{6}{9}=1\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
