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\( \left|\begin{array}{c}
a,\ b,\ c>0\\
a+b+c=1\end{array}\right|\Longrightarrow\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\ge\sqrt{\frac{3}{2}}. \)

Cezar Lupu, lista scurta 2005

Aplicand inegalitatea lui Holder avem :

\( \left( \frac {a}{\sqrt {b + c}} + \frac {b}{\sqrt {c + a}} + \frac {c}{\sqrt {a + b}}\right)^{2}\left( a(b + c) + b(c + a) + c(a + b) \right) \geq (a + b + c)^{3} \)

Deci

\( \sum\frac {a}{\sqrt {b + c}}\geq \sqrt {\frac {\left( \sum a \right)^{3}}{2\sum ab}}\geq \sqrt {\frac {3\left( \sum a \right)^{3}}{2\left( \sum a \right)^{2}}} = \sqrt {\frac {3}{2}\left( a + b + c \right)} \)

\( LHS=\sum\frac{1-(b+c)}{\sqrt{b+c}}=\sum\left\(\frac{1}{\sqrt{b+c}}-\sqrt{b+c}\right\) \)

Sa notam \( x=\sqrt{b+c},\ y=\sqrt{c+a},\ z=\sqrt{a+b} \)

\( \Longrightarrow LHS=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-(x+y+z) \)

Dar \( x^2+y^2+z^2=2 \) \( \Longrightarrow x+y+z\le \sqrt{6}\ (1) \)

Din inegalitatea \( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{9}{x+y+z} \)

\( \Longrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{9}{\sqrt{6}}=\frac{3\sqrt{6}}{2} \ (2) \)

Din \( (1) \) si \( (2) \) \( \Longrightarrow LHS\ge \frac{3\sqrt{6}}{2}-sqrt{6}=\sqrt{\frac{3}{2}} \)

Aplicam AM-GM :
\(

LHS = \sum_{cyc} \frac {x}{\sqrt {y + z}}= 2\sqrt {2}\cdot \sum_{cyc} \frac {x}{2\sqrt {2}\sqrt {y + z}}\ge 2\sqrt {2} \sum_{cyc} \frac {x}{2 + y + z} \ge \sqrt {2} \frac {(x + y + z)^2}{x + y + z + xy + yz + zx}\ge \sqrt{\frac{3}{2}} \)

O inegalitate putin mai tare :


Daca \( a,\ b,\ c>-1 \) a. i. \( a+b+c=1 \) atunci \( \sum\frac{a}{\sqrt{b+c}}\ge\frac{2}{3}\sum\frac{1}{\sqrt{1+a}}-\sqrt{\frac{3}{2}}. \)

Claudiu Mindrila wrote:\( a,\ b,\ c>0;\ a+b+c=1\Longrightarrow\sum\frac{a}{\sqrt{b+c}}\ge\sqrt{\frac{3}{2}} \)

Sa observam ca daca\( a\le b\le c\Longrightarrow\frac{1}{\sqrt{b+c}}\le\frac{1}{\sqrt{c+a}}\le\frac{1}{\sqrt{a+b}} \).

Atunci, cu inegalitatea lui Cebisev avem: \( \sum\frac{a}{\sqrt{b+c}}\ge\frac{1}{3}\left(\sum a\right)\left(\sum\frac{1}{\sqrt{1-a}}\right)=\frac{1}{3}\left(\sum\frac{1}{\sqrt{1-a}}\right). \quad (1) \).

Sa consideram acum functia \( f:\left[0,\ 1\right)\longrightarrow\mathbb{R},\: f\left(x\right)=\frac{1}{\sqrt{1-x}} \). Se demonstreaza usor(exercitiu) ca aceasta functie este convexa. Aplicand inegalitatea lui Jensen avem:

\( \frac{f\left(a\right)+f\left(b\right)+f\left(c\right)}{3}\ge f\left(\frac{a+b+c}{3}\right)\Longleftrightarrow f\left(a\right)+f\left(b\right)+f\left(c\right)\ge3f\left(\frac{1}{3}\right)=3\cdot\sqrt{\frac{3}{2}} \quad (2) \).

Din \( (1) \) si \( (2) \) avem concluzia problemei.

La inegalitatea aia cu \( >-1 \) ia \( b=c=-\frac{1}{2} \) si ai sa vezi ca nu se poate, asa ca nu e mai tare.

poate daca le pui pozitive merge.

Este intr-adevar mai slaba decat prima, greseala mea. Totusi, are cineva o demonstratie?
O sa postez cu alta ocazie o inegalitate mai tare decat prima, dar care se demonstreaza mai usor.

\( a,\ b,\ c>0\Longrightarrow\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\ge\left(a+b+c\right)\sqrt{\frac{a+b+c}{2\left(ab+bc+ca\right)}} \)

Cezar Lupu, OLM Constanta, 2006

Aplicam Cauchy-Schwartz :

\( LHS=\sum_{cyc}{\frac{a^2}{a{\sqrt{b+c}}}\ge\frac{(a+b+c)^2}{\sum{a\sqrt{b+c}}}\ge\frac{(a+b+c)^2}{\sqrt{2(a+b+c)(ab+bc+ca)}}=RHS \)