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Aratati ca daca \( a,\ b,\ c\ge0 \), cel mult unul din numere egal cu \( 0 \) , avem \( \frac{a^{2}+bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^{2}+ca}{\left(b+c\right)\left(b+a\right)}+\frac{c^{2}+ab}{\left(c+a\right)\left(c+b\right)}\ge\frac{a+b}{a+b+2c}+\frac{b+c}{b+c+2a}+\frac{c+a}{c+a+2b} \).

Inegalitatea este echivalenta cu

\( \sum\frac{(a-b)^2c}{(a+c)(b+c)(2a+b+c)(2b+a+c)}\ge 0 \)