Fie trei numere reale \( a,b,c\ge 0 \). Sa se demonstreze ca :
a) \( 8(a^3+b^3+c^3)^2\ge 9(a^2+bc)(b^2+ca)(c^2+ab) \)
b) \( (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)\ge (ab+bc+ca)^3 \)
Doua inegalitati in numere pozitive
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Doua inegalitati in numere pozitive
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Marius Mainea
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a) Inegalitatea este echivalenta cu \( 8\sum {a^6}+7\sum {a^3b^3}\ge 18a^2b^2c^2+9\sum {a^4bc} \) care este adevarata folosind AM-GM deoarece
\( a^6+a^6+b^3c^3\ge 3\sqrt[3]{a^{12}b^3c^3}=3a^4bc \)
\( \sum {a^3b^3}\ge 3a^2b^2c^2 \)
si
\( \sum {a^6}\ge \frac{1}{3}(a^3+b^3+c^3)^2\ge abc(a^3+b^3+c^3) \)
b) Aplicam inegalitatea lui Holder:
\( LHS=(a^2+b^2+ab)(ac+c^2+a^2)(c^2+bc+b^2)\ge (\sqrt[3]{a^2acc^2}+\sqrt[3]{b^2c^2bc}+\sqrt[3]{aba^2b^2})^3=RHS \)
\( a^6+a^6+b^3c^3\ge 3\sqrt[3]{a^{12}b^3c^3}=3a^4bc \)
\( \sum {a^3b^3}\ge 3a^2b^2c^2 \)
si
\( \sum {a^6}\ge \frac{1}{3}(a^3+b^3+c^3)^2\ge abc(a^3+b^3+c^3) \)
b) Aplicam inegalitatea lui Holder:
\( LHS=(a^2+b^2+ab)(ac+c^2+a^2)(c^2+bc+b^2)\ge (\sqrt[3]{a^2acc^2}+\sqrt[3]{b^2c^2bc}+\sqrt[3]{aba^2b^2})^3=RHS \)