[tex]\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. \\ \\ [/tex]
[tex]\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.[/tex]
