The minimum value of the sum of real numbers | vedclass

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Question

Since, $A M \geq G M$

$\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\left(\frac

Vedclass · Accepted Answer

$8$

Vedclass · Answer

Since, $A M \geq G M$

$\Rightarrow \frac{\frac{1}{a^5}+\frac{1}{a^4}+\frac{1}{a^3}+\frac{1}{a^3}+\frac{1}{a^3}+1+ a ^8+ a ^{10}}{8} \geq\left(\frac{1}{ a ^5} 	imes \frac{1}{ a ^4} 	imes \frac{1}{ a ^3} 	imes \frac{1}{ a ^3} 	imes \frac{1}{ a ^3} 	imes 1ight.$

$\Rightarrow \frac{\frac{1}{ a ^5}+\frac{1}{ a ^4}+\frac{1}{ a ^3}+\frac{1}{ a ^3}+\frac{1}{ a ^3}+1+ a ^8+ a ^{10}}{8} \geq(1)^{\frac{1}{8}}$

$\Rightarrow \frac{1}{ a ^5}+\frac{1}{ a ^4}+\frac{1}{ a ^3}+\frac{1}{ a ^3}+\frac{1}{ a ^3}+1+ a ^8+ a ^{10} \geq 8(1)^{\frac{1}{8}}$

$\Rightarrow \frac{1}{ a ^5}+\frac{1}{ a ^4}+\frac{1}{ a ^3}+\frac{1}{ a ^3}+\frac{1}{ a ^3}+1+ a ^8+ a ^{10} \geq 8$

So, the minimum value is $8$

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

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