Inegalitate conditionata cu a+b+c=1

[Claudiu Mindrila]

Fie \( a,\ b,\ c\in\left(0,\ \infty\right) \) cu \( a+b+c=1 \). Demonstrati ca \( \sqrt{\frac{ab}{ab+c}}+\sqrt{\frac{bc}{bc+a}}+\sqrt{\frac{ca}{ca+b}}\le\frac{3}{2} \).

Marin Chirciu, R.M.T. 2/2009

[alex2008]

Vezi aici .

[DrAGos Calinescu]

\( \sum \sqrt {\frac{ab}{ab+c}}=\sum\sqrt{\frac{ab}{ab+1-a-b}}=\sum\sqrt{\frac{ab}{(1-a)(1-b)}} \)
Notam \( x=\frac{a}{1-a} \) si analoagele.
Scoatem \( a,b,c \) in functie de \( x,y,z \) si din conditia initiala obtinem \( xy+yz+zx+2xyz=1 \) deci putem considera \( \sqrt{xy}=\cos C \)si analoagele.
Ramane de demonstrat binecunoscuta inegalitate dintr-un triunghi:
\( \cos A+\cos B+\cos C\le\frac{2}{3} \) (Jensen pt functii concave)

[Claudiu Mindrila]

Luand \( a=\frac{x}{x+y+z},\ b=\frac{y}{x+y+z},\ c=\frac{z}{x+y+z} \) cu \( x,\ y,\ z>0 \) avem:

\( \sum_{cyc}\sqrt{\frac{ab}{ab+c}}=\sum_{cyc}\sqrt{\frac{\frac{xy}{\left(x+y+z\right)^{2}}}{\frac{xy}{\left(x+y+z\right)^{2}}+\frac{z}{x+y+z}}}=\sum_{cyc}\sqrt{\frac{xy}{xy+z\left(x+y+z\right)}}=\sum_{cyc}\sqrt{\frac{xy}{\left(z+x\right)\left(z+y\right)}}\le\frac{1}{2}\left(\sum_{cyc}\frac{x}{x+y}+\sum_{cyc}\frac{y}{x+y}\right)=\frac{3}{2} \) , conform AM-GM.