Question

The minimum value of the sum of real numbers ${a^{ - 5}}, {a^{ - 4}}, 3{a^{ - 3}}, 1, {a^8}, {a^{10}}$ with $a > 0$ is
A) 6
B) 7
C) 8
D) 9

Answer 1

Hint: The inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list i.e.,$AM \geqslant GM$.
For two positive real numbers $x$ and $y$, AM is $\dfrac{{x + y}}{2}$ and GM is ${\left( {xy} \right)^{\dfrac{1}{2}}}$.

Complete step-by-step answer:
Given real numbers are ${a^{ - 5}},{a^{ - 4}},3{a^{ - 3}},1,{a^8},{a^{10}}$ with $a > 0$ .It means ${a^{ - 5}},{a^{ - 4}},3{a^{ - 3}},1,{a^8},{a^{10}}$$ > 0$
The inequality of arithmetic and geometric means states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list i.e.,$AM \geqslant GM$.
AM of the given numbers= $\dfrac{{\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}}}}{8}$
GM of given numbers= ${\left( {\dfrac{1}{{{a^5}}} \times \dfrac{1}{{{a^4}}} \times \dfrac{1}{{{a^3}}} \times \dfrac{1}{{{a^3}}} \times \dfrac{1}{{{a^3}}} \times 1 \times {a^8} \times {a^{10}}} \right)^{\dfrac{1}{8}}}$
Now, putting the value in the relation:$AM \geqslant GM$
$\dfrac{{\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}}}}{8}$$ \geqslant $${\left( {\dfrac{1}{{{a^5}}} \times \dfrac{1}{{{a^4}}} \times \dfrac{1}{{{a^3}}} \times \dfrac{1}{{{a^3}}} \times \dfrac{1}{{{a^3}}} \times 1 \times {a^8} \times {a^{10}}} \right)^{\dfrac{1}{8}}}$
$ \Rightarrow $$\dfrac{{\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}}}}{8}$$ \geqslant $${\left( 1 \right)^{\dfrac{1}{8}}}$
$ \Rightarrow $ $\dfrac{{\dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}}}}{8}$$ \geqslant $$1$
$ \Rightarrow \dfrac{1}{{{a^5}}} + \dfrac{1}{{{a^4}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + \dfrac{1}{{{a^3}}} + 1 + {a^8} + {a^{10}} \geqslant 8$
$ \Rightarrow {a^{ - 5}} + {a^{ - 4}} + 3{a^{ - 3}} + 1 + {a^8} + {a^{10}} \geqslant 8$
Thus, the minimum value of the sum of real numbers ${a^{ - 5}},{a^{ - 4}},3{a^{ - 3}},1,{a^8},{a^{10}}$ with $a > 0$ is 8.

Hence, option (C) is the correct answer.

Note: In this question, we break the $3{a^{ - 3}}$ into ${a^{ - 3}},{a^{ - 3}},{a^{ - 3}}$ so that their multiplication gets easier and desired result can be obtained without any difficulty.
${a^{-m}}=\dfrac{1}{{a^m}}$