1. Consider the expression $$\frac{(a^2+a+1)(b^2+b+1)(c^2+c+1)(d^2+d+1)(e^2+e+1)}{abcde}$$, where a,b,c,d and e are positive numbers. The minimum value of the expression is
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By: anil on 05 May 2019 03.05 pm
The given expression can be written as $$frac{a^2+a+1}{a}*frac{b^2+b+1}{b}*frac{c^2+c+1}{c}*frac{d^2+d+1}{d}*frac{e^2+e+1}{e}$$.
$$frac{a^2+a+1}{a}=a+frac{1}{a}+1$$
We know that for positive values, AM $$geq$$ GM.
$$frac{1+frac{1}{a}}{2} geq sqrt{a*frac{1}{a}}$$
$$a+frac{1}{a} geq 2$$
The least value that $$a+frac{1}{a}$$ can take is $$2$$.
Therefore, the least value that the term $$a+frac{1}{a}+1$$ can take is $$3$$.
Similarly, the least value that the other terms can take is also $$3$$.
=> The least value of the given expression = $$3*3*3*3*3$$ = $$243$$.
Therefore, option E is the right answer.
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$$frac{a^2+a+1}{a}=a+frac{1}{a}+1$$
We know that for positive values, AM $$geq$$ GM.
$$frac{1+frac{1}{a}}{2} geq sqrt{a*frac{1}{a}}$$
$$a+frac{1}{a} geq 2$$
The least value that $$a+frac{1}{a}$$ can take is $$2$$.
Therefore, the least value that the term $$a+frac{1}{a}+1$$ can take is $$3$$.
Similarly, the least value that the other terms can take is also $$3$$.
=> The least value of the given expression = $$3*3*3*3*3$$ = $$243$$.
Therefore, option E is the right answer.