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Fie \( a,b,c>0 \). Demonstrati inegalitatea:
\( \frac{a^3}{a+1}+\frac{b^3}{b+1}+\frac{c^3}{c+1} \geq \frac{(a+b+c)^2-3(a+b+c)+9}{3}-\frac{(a+b+c+3)^2}{3(a+1)(b+1)(c+1)} \)
Claudiu Mindrila

Inegalitatea este echivalenta cu

\( \sum{\frac{a^3}{a+1}}+\frac{(a+b+c+3)^2}{3(a+1)(b+1)(c+1)}\ge \frac{(a+b+c)^2}{3}-(a+b+c)+3 \)

Folosind relatia \( (x+y+z)^2\ge3(xy+yz+zx) \) pentru x=a+1 , y=b+1, z=c+1 avem

\( LHS\ge \sum{\frac{a^3}{a+1}+\sum{\frac{1}{a+1}}=\sum{\frac{a^3+1}{a+1}}=\sum{(a^2-a+1)}\ge\frac{(a+b+c)^2}{3}-\sum{a}+3=RHS \)