The minimum value of the sum of real numbers | vedclass

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(C) Given the expression $S = \frac{1}{a^5} + \frac{1}{a^4} + \frac{3}{a^3} + 1 + a^8 + a^{10}$.We can write this as $S = \frac{1}{a^5} + \frac{1}{a^4

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Given the expression $S = \frac{1}{a^5} + \frac{1}{a^4} + \frac{3}{a^3} + 1 + a^8 + a^{10}$.We can write this as $S = \frac{1}{a^5} + \frac{1}{a^4} + \frac{1}{a^3} + \frac{1}{a^3} + \frac{1}{a^3} + 1 + a^8 + a^{10}$.Applying the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for these $8$ positive terms:$\frac{\frac{1}{a^5} + \frac{1}{a^4} + \frac{1}{a^3} + \frac{1}{a^3} + \frac{1}{a^3} + 1 + a^8 + a^{10}}{8} \geq \sqrt[8]{\frac{1}{a^5} \cdot \frac{1}{a^4} \cdot \frac{1}{a^3} \cdot \frac{1}{a^3} \cdot \frac{1}{a^3} \cdot 1 \cdot a^8 \cdot a^{10}}$.The product of the terms is $\frac{1}{a^{5+4+3+3+3}} \cdot a^{8+10} = \frac{1}{a^{18}} \cdot a^{18} = 1$.Thus,$\frac{S}{8} \geq \sqrt[8]{1} = 1$.Therefore,$S \geq 8$.The minimum value is $8$.

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

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