Question

Daca a,b,c-lungimile laturilor unui triunghi ,aratati ca are loc inegalitatea :
$\sqrt{ \frac{a}{-a+b+c} } + \sqrt{ \frac{b}{a-b+c} } + \sqrt{ \frac{c}{a+b-c} } \geq 3$
Filip,e urgenta problema !:))(te rog)

Answer 1

$Am~luat~maxim~pe~aceasta~problema. \\ \\ Din~inegalitatea~mediilor~(m_g \geq m_h)~avem: \\ \\ \sqrt{x}= \sqrt{x \cdot 1} \geq \frac{2x}{x+1}. \\ \\ Avem~deci:~ \sqrt{ \frac{a}{-a+b+c} } \geq \frac{2 \cdot \frac{a}{-a+b+c} }{ \frac{a}{-a+b+c}+1 }= \frac{2a}{b+c}. \\ \\ Membrul~stang \geq 2( \frac{a}{b+c}+ \frac{b}{a+c}+ \frac{c}{a+b}). \\ \\ Si~din~inegalitatea~lui~Nesbitt~avem~ \frac{a}{b+c}+ \frac{b}{a+c}+ \frac{c}{a+b} \geq \frac{3}{2} , \\ \\ de~unde~rezulta~concluzia.$

$La~momentul~respectiv~nu~am~precizat~ca~aceea~era \\ \\ inegalitatea~lui~Nesbitt,~ci~pur~si~simplu~am~demonstrat-o: \\ \\ \frac{a}{b+c}+ \frac{b}{a+c}+ \frac{c}{a+b}= \frac{a^2}{ab+ac}+ \frac{b^2}{ab+bc}+ \frac{c^2}{ac+bc} \geq ~(Cauchy- \\ \\ Buniakovski-Schwartz) \geq \frac{(a+b+c)^2}{2ab+2bc+2ac}= \frac{a^2+b^2+c^2+2ab+2bc+2ac}{2ab+2bc+2ac} \\ \\ = \frac{a^2+b^2+c^2}{2(ab+bc+ac)}+1 \geq \frac{ab+bc+ac}{2(ab+bc+ac)} +1= \frac{3}{2}.$

$Deci~Membrul~Stang \geq 2( \frac{a}{b+c}+ \frac{b}{a+c}+ \frac{c}{a+b}) \geq 2 \cdot \frac{3}{2}=3. \\ \\ *Am~folosit~si~inegalitatea~x^2+y^2+z^2 \geq xy+yz+xz. \\ \\ Te~rog,~fara~reprosuri~referitoare~la~inegalitea~C-B-S. \\ \\ Aceasta~este~cam~cea~mai~importanta~inegalitate,~si~trebuie \\ \\ cunoscuta~neaparat~pentru~olimpiada!~Felicitari~pentru~ \\ \\ calificare~si~mult~succes!$