An easy one from RMM



  • if a,b,c(0,)a,b,c\in(0,\infty) such that
    3(a+b)(b+c)(c+a)8a3+b3+c383(a+b)(b+c)(c+a)\ge\frac{8}{\sqrt[8]{a^3+b^3+c^3}}
    Prove that
    a+b+c93a+b+c\ge\sqrt[3]{9}



  • On a :3(a+b)(b+c)(c+a)8a3+b3+c383(a+b)(b+c)(c+a) \geq \frac{8}{\sqrt[8]{a^3+b^3+c^3}}
    (a+b+c)3(a3+b3+c3)8a3+b3+c38\Leftrightarrow (a+b+c)^3 - (a^3+b^3+c^3) \geq \frac{8}{\sqrt[8]{a^3+b^3+c^3}}
    D'apres I.A.GI.A.G:
    (a+b+c)38a3+b3+c38+(a3+b3+c3)9a3+b3+c3(a3+b3+c38)89\Leftrightarrow (a+b+c)^3 \geq \frac{8}{\sqrt[8]{a^3+b^3+c^3}}+(a^3+b^3+c^3) \ge 9\sqrt[9]{\frac{a^3+b^3+c^3}{(\sqrt[8]{a^3+b^3+c^3})^8}}
    (a+b+c)39\Leftrightarrow (a+b+c)^3 \ge 9
    a+b+c93\Leftrightarrow a+b+c \ge \sqrt[3]{9}



  • @Reda-Mouqed perfect!


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