If a,b,and c are non-zero real numbers,then t | vedclass

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Question

(B) The given expression is $\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}}\ri

Vedclass · Accepted Answer

$351$

Vedclass · Answer

(B) The given expression is $\left(\frac{a^{4}+a^{2}+1}{a^{2}}ight)\left(\frac{b^{4}+7 b^{2}+1}{b^{2}}ight)\left(\frac{c^{4}+11 c^{2}+1}{c^{2}}ight)$.This can be rewritten as $\left(a^{2}+\frac{1}{a^{2}}+1ight)\left(b^{2}+\frac{1}{b^{2}}+7ight)\left(c^{2}+\frac{1}{c^{2}}+11ight)$.Using the identity $x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2$,we get:$\left[(a-\frac{1}{a})^{2}+3ight]\left[(b-\frac{1}{b})^{2}+9ight]\left[(c-\frac{1}{c})^{2}+13ight]$.The minimum value of $(x - \frac{1}{x})^2$ is $0$,which occurs when $x^2 = 1$.Thus,the minimum value of the expression is $3 	imes 9 	imes 13 = 351$.

If $a$,$b$,and $c$ are non-zero real numbers,then the minimum value of the expression $\left( \frac{(a^4 + a^2 + 1)(b^4 + 7b^2 + 1)(c^4 + 11c^2 + 1)}{a^2 b^2 c^2} \right)$ is

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