Daca a,b,c sunt laturile unui triunghi , atunci :
\( \frac{a+b}{c^3+abc}+\frac{b+c}{a^3+abc}+\frac{c+a}{b^3+abc}\ge\frac{1}{2Rr}. \)
Inegalitate geometrica 2
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Marius Mainea
- Gauss
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Avem :
\( 2Rr = 2\frac {abc}{4S}\cdot \frac {S}{p} = \frac {abc}{a + b + c} \) , deci inegalitatea este echivalenta cu :
\( \frac {a + b}{c^2 + abc} + \frac {b + c}{a^3 + abc} + \frac {c + a}{b^3 + abc} - \frac {a + b + c}{abc}\geq0\Leftrightarrow \frac {(a^2 - b^2)^2 + (a^2 - c^2)^2 + (b^2 - c^2)^2}{2(b^2 + ac)(a^2 + bc)(c^2 + ab)}\geq 0 \)
\( 2Rr = 2\frac {abc}{4S}\cdot \frac {S}{p} = \frac {abc}{a + b + c} \) , deci inegalitatea este echivalenta cu :
\( \frac {a + b}{c^2 + abc} + \frac {b + c}{a^3 + abc} + \frac {c + a}{b^3 + abc} - \frac {a + b + c}{abc}\geq0\Leftrightarrow \frac {(a^2 - b^2)^2 + (a^2 - c^2)^2 + (b^2 - c^2)^2}{2(b^2 + ac)(a^2 + bc)(c^2 + ab)}\geq 0 \)
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