Fie \( a,b,c \) lungimile laturilor unui triunghi. Demonstrati: \( \frac{(-a+b+c)^3}{a^2}+\frac{(a-b+c)^3}{b^2}+\frac{(a+b-c)^3}{c^2} \geq a+b+c. \)
Gheorghe Stoica
Inegalitate cu laturile unui triunghi
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Claudiu Mindrila
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Inegalitate cu laturile unui triunghi
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Marius Mainea
- Gauss
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Tripletele \( (b+c-a,c+a-b,a+b-c) \) si \( (\frac{b+c-a}{a},\frac{c+a-b}{b},\frac{a+b-c}{c}) \) sunt la fel orientate si aplicand inegalitatea rearanjamentelor , inegalitatea CBS si inegalitatea mediilor avem :
\( LHS\ge \frac{1}{3}\cdot\sum{(\frac{b+c-a}{2})^2}\cdot\sum{(b+c-a)}\ge \frac{1}{9}\cdot(\sum{\frac{b+c-a}{a}})^2\cdot(a+b+c)=\frac{1}{9}\cdot(\sum(\frac{a}{b}+\frac{b}{a})-3)^2\cdot(a+b+c)\ge \frac{1}{9}\cdot(2+2+2-3)^2(a+b+c)=RHS \)
\( LHS\ge \frac{1}{3}\cdot\sum{(\frac{b+c-a}{2})^2}\cdot\sum{(b+c-a)}\ge \frac{1}{9}\cdot(\sum{\frac{b+c-a}{a}})^2\cdot(a+b+c)=\frac{1}{9}\cdot(\sum(\frac{a}{b}+\frac{b}{a})-3)^2\cdot(a+b+c)\ge \frac{1}{9}\cdot(2+2+2-3)^2(a+b+c)=RHS \)