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G.M. 3/2008
Posted: Sun May 11, 2008 1:50 pm
by BogdanCNFB
Sa se arate ca \( \frac{4a^2+2a +1}{a}-\frac{12a}{4a^2+2a+1}\ge 4,\forall a>0 \).
Posted: Tue May 13, 2008 1:44 pm
by Marius Dragoi
\( \frac {4a^2+2a+1}{a} - \frac {12a}{4a^2+2a+1} \geq 4 \) \( \Leftrightarrow \) \( {(4a^2-1)}^2 \geq 0 \).