MCQ
If $a$ , $b$ , $c$ are non zero real numbers, then minimum value of the expression $\left( {\frac{{\left( {{a^4} + {a^2} + 1} \right)\left( {{b^4} + 7{b^2} + 1} \right)\left( {{c^4} + 11{c^2} + 1} \right)}}{{{a^2}{b^2}{c^2}}}} \right)$ is
  • A
    $315$
  • $351$
  • C
    $415$
  • D
    $451$

Answer

Correct option: B.
$351$
b
$\left(\frac{a^{4}+a^{2}+1}{a^{2}}\right)\left(\frac{b^{4}+7 b^{2}+1}{b^{2}}\right)\left(\frac{c^{4}+11 c^{2}+1}{c^{2}}\right)$

$\Rightarrow\left(a^{2}+\frac{1}{a^{2}}+1\right)\left(b^{2}+\frac{1}{b^{2}}+7\right)\left(c^{2}+\frac{1}{c^{2}}+11\right)$

$\left[\left(a-\frac{1}{a}\right)^{2}+3\right]\left[\left(b-\frac{1}{b}\right)^{2}+9\right]\left[\left(c-\frac{1}{c}\right)^{2}+13\right]$

clearly minimum value occurs when $a^{2}=b^{2}=c^{2}=1$ and minimum value

$=3 \times 9 \times 13=351$

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