Áp dụng BĐT Cô-si:
\(a^2+3\ge2\sqrt{3a^2}=2\sqrt{3}a\)
Tương tự: \(b^2+3\ge2\sqrt{3}b\) ; \(c^2+3\ge2\sqrt{3}c\)
Cộng vế: \(a^2+b^2+c^2+9\ge2\sqrt{3}\left(a+b+c\right)\)
\(\Rightarrow a+b+c\le\dfrac{a^2+b^2+c^2+9}{2\sqrt{3}}=\dfrac{9+9}{2\sqrt{3}}=3\sqrt{3}\)
\(\Rightarrow-\left(a+b+c\right)\ge-3\sqrt{3}\)
Tiếp tục áp dụng BĐT Cô-si:
\(\dfrac{a^4}{b+2}+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(b+2\right)\ge2\sqrt{\dfrac{9a^4\left(b+2\right)}{\left(b+2\right)\left(2+\sqrt{3}\right)^2}}=\dfrac{6a^2}{2+\sqrt{3}}\)
Tương tự:
\(\dfrac{b^4}{c+2}+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(c+2\right)\ge\dfrac{6b^2}{2+\sqrt{3}}\)
\(\dfrac{c^4}{a+2}+\dfrac{9}{\left(2+\sqrt{3}\right)}\left(a+2\right)\ge\dfrac{6c^2}{2+\sqrt{3}}\)
Cộng vế:
\(P+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(a+b+c+6\right)\ge\dfrac{6}{2+\sqrt{3}}\left(a^2+b^2+c^2\right)=\dfrac{54}{2+\sqrt{3}}\)
\(\Rightarrow P\ge\dfrac{54}{2+\sqrt{3}}-\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(a+b+c+6\right)\ge\dfrac{54}{2+\sqrt{3}}-\dfrac{9}{\left(2+\sqrt{3}\right)^2}.\left(3\sqrt{3}+6\right)\)
\(\Rightarrow P\ge\dfrac{27}{2+\sqrt{3}}=27\left(2-\sqrt{3}\right)\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
