Question

$Pentru~orice~numere~reale~nenule~a,b,c,aratati ~ca: \\ \sqrt{ \frac{a+c}{b} } + \sqrt{\frac{a+b}{c} }+ \sqrt{ \frac{b+c}{a} }\ \textgreater \ 2$

Answer 1

$\displaystyle Din~inegalitatea~dintre~media~geometrica~si~media ~armonica,~\\ \\avem:~ \sqrt{\frac{b+c}{a}}=\sqrt{\frac{b+c}{a} \cdot 1} \ge \frac{2 \cdot \frac{b+c}{a} \cdot 1}{\frac{b+c}{a}+1}= \frac{2(b+c)}{a+b+c}. \\ \\ Egalitate \Leftrightarrow a=b+c. \\ \\ Scriind~si~relatiile~similare,~si~adunandu-le,~obtinem: \\ \\ \sum \sqrt{\frac{b+c}{a}} \ge \sum \frac{2(b+c)}{a+b+c}=2.$

$\displaystyle Egalitatea~ar~avea~loc~daca~a=b+c,~b=a+c,~c=a+b,~ceea~ce \\ \\ ar~implica~a=b=c=0,~imposibil!~Deci~inegaliatea~este~stricta.$