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Fie \( a, b, c > 0 \) cu \( a+b+c=1 \). Sa se arate ca:
1) \( \sum \frac{\sqrt{a+bc}}{\sqrt{bc} + \sqrt{a+bc}} \ge 2 \).
2) \( \sum \frac{\sqrt{a+bc}}{\sqrt{a} + \sqrt{a+bc}} \ge \frac{6}{2+\sqrt{3}} \).
3) \( \sum \frac{\sqrt{a+bc}}{\sqrt{a} + \sqrt{bc}} \ge \frac{6}{1+\sqrt{3}} \).
Topicul exista si pe mathlinks. Poate aici lumea are mai multe idei

Notez \( E_{1} =\sum \sqrt{\frac{a}{(a+b)(a+c)}}\leq \frac{3\sqrt{3}}{2} \) cu conditia \( a+b+c=1 \)

\( E_{2}=\sum \sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{3}{2} \)

Voi demonstra inegalitatea :\( E_{1}\leq \frac{3\sqrt{3}}{2} \)

LHS \( \leq 3\sqrt{\frac{1}{3}\sum \frac{a}{(a+b)(a+c)}}\leq \frac{3\sqrt{3}}{2} \Leftrightarrow \)
\( \sum \frac{a}{(a+b)(a+c)}\leq \frac{9}{4} \Leftrightarrow \)
\( 8(ab+bc+ca)(a+b+c)\leq 9(a+b)(b+c)(c+a) \Leftrightarrow \)
\( 8abc \leq (a+b)(b+c)(c+a) \) evident ma-mg

In continuare voi demonstra \( E_{2}\leq \frac{3}{2} \)
\( E_{1}\leq \frac{1}{2} \sum \left(\frac{b}{a+b}+\frac{c}{a+c}\right) =\frac{3}{2} \)

Astfel putem demonstra inegalitatile 1,2,3 cu rezultatele de mai sus:

1)

\( LHS\geq\frac{9}{E_{1}+3}\geq \frac{9}{\frac{3}{2}+3}=2 \)

2)
\( LHS\geq \frac{9}{E_{2}+3}\geq \frac{9}{\frac{3\sqrt{3}}{2}+3}=\frac{6}{2+\sqrt{3}} \)

3)
\( LHS\geq\frac{9}{E_{1}+E_{2}} \geq \frac{9}{\frac{3\sqrt{3}}{2}+\frac{3}{2}}=\frac{6}{\sqrt{3}+1} \)