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Given $a,b,c > 0,a^2=bc$ and $a+b+c=abc$,the least possible value of $a^2$ is

$\begin{array}{1 1}(A)\;1&(B)\;2\\(C)\;3&(D)\;5\end{array} $

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$b$ and $c$ are roots of $x^2-(a^3-a)x+a^2=0$
As b and c are real
$(a^3-a)^2-4a^2\geq 0$
$\Rightarrow a^2(a^2+1)(a^2-3)\geq 0$
$\Rightarrow a^2 \geq 3$
Hence (C) is the correct answer.
answered Apr 16, 2014 by sreemathi.v
 

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