Inegalitate "nice"

[Claudiu Mindrila]

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

[Marius Mainea]

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

[Claudiu Mindrila]

Aveti dreptate. Am modificat.

[Marius Dragoi]

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

[Claudiu Mindrila]

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