G.M. 3/2008

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BogdanCNFB
Thales
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Location: Craiova

G.M. 3/2008

Post by BogdanCNFB »

Sa se arate ca \( \frac{4a^2+2a +1}{a}-\frac{12a}{4a^2+2a+1}\ge 4,\forall a>0 \).
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Marius Dragoi
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Post by Marius Dragoi »

\( \frac {4a^2+2a+1}{a} - \frac {12a}{4a^2+2a+1} \geq 4 \) \( \Leftrightarrow \) \( {(4a^2-1)}^2 \geq 0 \).
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
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