The minimum value of the sum of real numbers $a^{- 5}, a^{- 4}, 3 a^{- 3}, 1, a^{8}$ and $a^{10}$ with $a > 0$ is

6

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7

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8

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9

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Solution

The correct option is C $8$
Since, AM $\geq$ GM

$\Rightarrow \frac{\frac{1}{a^{5}} + \frac{1}{a^{4}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + 1 + a^{8} + a^{10}}{8} \geq {(\frac{1}{a^{5}} \times \frac{1}{a^{4}} \times \frac{1}{a^{3}} \times \frac{1}{a^{3}} \times \frac{1}{a^{3}} \times 1 \times a^{8} \times a^{10})}^{\frac{1}{8}}$

$\Rightarrow \frac{\frac{1}{a^{5}} + \frac{1}{a^{4}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + 1 + a^{8} + a^{10}}{8} \geq {(1)}^{\frac{1}{8}}$

$\Rightarrow \frac{1}{a^{5}} + \frac{1}{a^{4}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + 1 + a^{8} + a^{10} \geq 8 {(1)}^{\frac{1}{8}}$

$\Rightarrow \frac{1}{a^{5}} + \frac{1}{a^{4}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + \frac{1}{a^{3}} + 1 + a^{8} + a^{10} \geq 8$

So, the minimum value is $8$

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