mateforum.ro

Fie \( a, b, c>0 \). Sa se arate ca:
\( a^3+b^3+c^3\ge\frac{a^2b^2(a+b)}{a^2+b^2}+\frac{b^2c^2(b+c)}{b^2+c^2}+\frac{a^2c^2(a+c)}{a^2+c^2} \).
I.V.Maftei, Marius Radulescu
(rev. Arhimede)

Avem:

\( \sum\frac{a^{2}b^{2}\left(a+b\right)}{a^{2}+b^{2}}\le\sum\frac{a^{2}b^{2}\left(a+b\right)}{2ab}=\frac{1}{2}\sum ab\left(a+b\right)\le\frac{1}{2}\sum\left(a^{3}+b^{3}\right)=\sum a^{3} \),

unde mai sus am folosit inegalitatea mediilor si faptul ca \( \sum\left(a-b\right)^{2}\left(a+b\right)\ge0\Longleftrightarrow\sum a^{3}\ge\sum ab\left(a+b\right) \). Egalitatea are loc cand \( a=b=c \).