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A.M Greater than or Equal to G.M

If a 1+2 a 2+...

Question

If $a_{1} + 2 a_{2} + 3 a_{3} = 1$ , where $a_{1}, a_{2} & a_{3}$ are all positive numbers; then what is the minimum value of $2^{a 1} + 4^{a 2} + 8^{a 3}$ ?

4^1/3

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3*2^1/3

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2^1/3

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3*4^1/3

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Solution

The correct option is D
3*4^1/3

AM $= \frac{(2^{a_{1}} + 4^{a_{2}} + 8^{a_{3}})}{3}$ and GM $= (2^{a_{1}} \times 4^{a_{2}} \times 8^{a_{3}})^{(\frac{1}{3})}$

Using condition AM $\geq$ GM, we get $\frac{(2^{a_{1}} + 4^{a_{2}} + 8^{a_{3}})}{3} \geq (2^{a_{1}} \times 4^{a_{2}} \times 8^{a_{3}})^{(\frac{1}{3})}$

$2^{a_{1}} + 4^{a_{2}} + 8^{a_{3}} \geq 3 (2^{a_{1}} \times 2^{2 a_{2}} \times 2^{3 a_{3}})^{(\frac{1}{3})}$

$\geq 3 (2^{a_{1} + 2 a_{2} + 3 a_{3}})^{(\frac{1}{3})}$

But $a_{1} + 2 a_{2} + 3 a_{3} = 1$

Answer = $(27 \times 2)^{\frac{1}{3}}$

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