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Fie \( a,\ b,\ c>0 \). Sa se arate ca: \( a^{3}+b^{3}+c^{3}\ge\frac{a^{2}b^{2}\left(a+b\right)}{a^{2}+b^{2}}+\frac{b^{2}c^{2}\left(b+c\right)}{b^{2}+c^{2}}+\frac{c^{2}a^{2}\left(c+a\right)}{c^{2}+a^{2}} \).

I. V. Maftei, Marius Radulescu

Avem \( a^2+b^2\ge 2ab \) si analoagele.

\( \Longrightarrow RHS\le \sum\frac{a^2b^2(a+b)}{2ab}=\sum\frac{ab(a+b)}{2} \)

Ramane sa aratam ca \( 2(a^3+b^3+c^3)\ge ab(a+b)+bc(b+c)+ca(c+a) \)

Dar \( a^3+b^3\ge ab(a+b)=a^2b+b^2a \Longleftrightarrow (a-b)^2(a+b)\ge 0 \)

Adunand si analoagele obtinem cerinta.