Ta có: \(2.2.\sqrt{x^2+3}\le x^2+3+4=x^2+7\Leftrightarrow\sqrt{x^2+3}\le\frac{x^2+7}{4}\) (đẳng thức xảy ra khi x = 1.)
Áp dụng BĐT trên ta có:
\(P\ge4\left(\frac{a^3}{b^2+7}+\frac{b^3}{c^2+7}+\frac{c^3}{a^2+7}\right)=4.\left(\frac{a^4}{ab^2+7a}+\frac{b^4}{bc^2+7b}+\frac{c^4}{ca^2+7c}\right)\ge4.\frac{\left(a^2+b^2+c^2\right)^2}{ab^2+bc^2+ca^2+7\left(a+b+c\right)}\)
( Theo BĐT Schwarz)
Áp dụng BĐT Bunhiacopxki với 3 số ta có:
\(\left(ab^2+bc^2+ca^2\right)^2=\left(b.ab+c.bc+a.ca\right)^2\le\left(a^2+b^2+c^2\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\le\left(a^2+b^2+c^2\right)\frac{\left(a^2+b^2+c^2\right)^2}{3}=\frac{\left(a^2+b^2+c^2\right)^3}{3}=\frac{3^3}{3}=9\Rightarrow ab^2+bc^2+ca^2\le3\)
Ta có: \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)=9\Rightarrow a+b+c\le3\)
Do đó:
\(P\ge4.\frac{\left(a^2+b^2+c^2\right)^2}{ab^2+bc^2+ca^2+7\left(a+b+c\right)}\ge\frac{4.3^2}{3+7.3}=\frac{3}{2}\)
Xảy ra đẳng thức khi a = b = c = 1.
Vậy min \(P=\frac{3}{2}\) khi a = b = c = 1.
