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# If $\;px^2+\large\frac{q}{x}\;\geq\;r\;$ for every +ve x $\;(p>0 , q>0)\;,\;$ then $\;27pq^2\;$ can not be less than

$(a)\;r^3\qquad(b)\;4r^3\qquad(c)\;8r^3\qquad(d)\;4r^2$

Answer : (b) $\;4r^3$
Explanation : $\;AM\;\geq\;GM$
$\frac{1}{3}\;(px^2+\large\frac{q}{2x}+\large\frac{q}{2x})\;\geq\;(px^2.\large\frac{q}{2x}.\large\frac{q}{2x})$
$\large\frac{1}{3}(px^2+\large\frac{q}{x})\;\geq\;(\large\frac{pq^2}{4})^{\frac{1}{3}}$
Least value of $\;px^2+\frac{q}{x}\;$ is r
$3(\large\frac{pq^2}{4})^{\frac{1}{3}}\;\geq\;r$
$27pq^2\;\geq\;4r^3.$