Fie \( a,\ b,\ c>0 \). Aratati ca \( 1+\frac{1}{6abc}>\frac{3}{a+b+c} \).
Claudiu Mindrila, Revista de Matematica din Timisoara nr. 3/2010
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Claudiu Mindrila
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Andi Brojbeanu
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Din ineg. mediilor avem : \( abc\le\frac{(a+b+c)^3}{3}\Rightarrow 1+\frac{1}{6abc}\ge 1+\frac{1}{\frac{9}{2}(a+b+c)^3}=1+\frac{9}{2(a+b+c)^3} \).
Notand \( a+b+c=x \), obtinem ca \( LHS\ge 1+\frac{9}{2x^3}=\frac{2x^3+9}{2x^3}=\frac{(x^3+x^3+{8})+1}{2x^3}\ge \frac{3\sqrt[3]{x^3\cdot x^3\cdot 2^3}+1}{2x^3}=\frac{3\cdot 2x^2+1}{2x^3}>\frac{6x^2}{2x^3}=\frac{3}{x}=RHS \).
Notand \( a+b+c=x \), obtinem ca \( LHS\ge 1+\frac{9}{2x^3}=\frac{2x^3+9}{2x^3}=\frac{(x^3+x^3+{8})+1}{2x^3}\ge \frac{3\sqrt[3]{x^3\cdot x^3\cdot 2^3}+1}{2x^3}=\frac{3\cdot 2x^2+1}{2x^3}>\frac{6x^2}{2x^3}=\frac{3}{x}=RHS \).
Andi Brojbeanu
profesor, Liceul Teoretic "Lucian Blaga", Cluj-Napoca
profesor, Liceul Teoretic "Lucian Blaga", Cluj-Napoca