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Fie \( a,b,c \in \mathbb{R}^*_+ \). Demonstrati inegalitatea:
\( \frac{\sqrt{a}}{(a+b)(a+c)}+\frac{\sqrt{b}}{(b+a)(b+c)}+\frac{\sqrt{c}}{(c+a)(c+b} \leq \frac{3}{4\sqrt{abc}} \).

Claudiu Mindrila wrote:Fie \( a,b,c \in \mathbb{R}^*_+ \). Demonstrati inegalitatea:
\( \frac{\sqrt{a}}{(a+b)(a+c)}+\frac{\sqrt{b}}{(b+a)(b+c)}+\frac{\sqrt{c}}{(c+a)(c+b} \geq \frac{3}{4\sqrt{abc}} \).

Poate \( \frac{\sqrt{a}}{(a+b)(a+c)}+\frac{\sqrt{b}}{(b+a)(b+c)}+\frac{\sqrt{c}}{(c+a)(c+b} \leq \frac{3}{4\sqrt{abc}} \)

Aveti dreptate. Am modificat.

\( \sum_{} {\frac {\sqrt {a}}{(a+b)(a+c)} \) \( \leq \) \( \sum_{} {\frac {\sqrt {a}}{4 \sqrt {a^2bc}} \) \( = \) \( \frac {3}{4 \sqrt {abc}} \)

Asta a fost si solutia mea Marius. Insa daca vrei o inegalitate draguta, incearca aici .