Din nou o inegalitate "nice"
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Claudiu Mindrila
- Fermat
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Din nou o inegalitate "nice"
Fie \( a,b,c \) numere nenegative. Demonstrati ca \( \sqrt{\frac{a^3}{a^3+(b+c)^3}}+\sqrt{\frac{b^3}{b^3+(c+a)^3}}+\sqrt{\frac{c^3}{c^3+(a+b)^3}} \geq 1 \)
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
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Marius Mainea
- Gauss
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\( \sqrt{1+x^3}=\sqrt{(1+x)(1-x+x^2)}\le \frac{(1+x)+(1-x+x^2)}{2}=1+\frac{x^2}{2} \) pentru \( x\ge 0 \)
Asadar \( \sqrt{\frac{a^3}{a^3+(b+c)^3}}=\frac{1}{\sqrt{1+(\frac{b+c}{a})^3}}\ge\frac{1}{1+\frac{1}{2}(\frac{b+c}{a})^2}\ge \frac{1}{1+\frac{b^2+c^2}{a^2}}=\frac{a^2}{a^2+b^2+c^2} \)
Se procedeaza analog cu ceilalti doi termeni si se aduna.
Asadar \( \sqrt{\frac{a^3}{a^3+(b+c)^3}}=\frac{1}{\sqrt{1+(\frac{b+c}{a})^3}}\ge\frac{1}{1+\frac{1}{2}(\frac{b+c}{a})^2}\ge \frac{1}{1+\frac{b^2+c^2}{a^2}}=\frac{a^2}{a^2+b^2+c^2} \)
Se procedeaza analog cu ceilalti doi termeni si se aduna.