\(\frac{a^3}{1+b}+\frac{1+b}{4}+\frac{1}{2}\ge\frac{3}{2}a\) ; \(\frac{b^3}{1+c}+\frac{1+c}{4}+\frac{1}{2}\ge\frac{3}{2}b\) ; \(\frac{c^3}{1+a}+\frac{1+a}{4}+\frac{1}{2}\ge\frac{3}{2}c\)
Cộng vế với vế:
\(P+\frac{1}{4}\left(a+b+c\right)+\frac{9}{4}\ge\frac{3}{2}\left(a+b+c\right)\)
\(\Rightarrow P\ge\frac{5}{4}\left(a+b+c\right)-\frac{9}{4}\ge\frac{5}{4}\sqrt{3\left(ab+bc+ca\right)}-\frac{9}{4}\ge\frac{5}{4}\sqrt{9}-\frac{9}{4}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)