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If $G,O,E$ and $L$ are positive real numbers such that $$\log_{10}(G \cdot L) + \log_{10}(G \cdot E) =3$$ $$\log_{10}(E \cdot L) + \log_{10}(O \cdot E) =4$$ $$\log_{10}(G \cdot O) + \log_{10}(L \cdot O) =5$$

If the minimum value of $3G +2L+ 2O+E$ is $2^\lambda \cdot 3^\mu \cdot 5^\nu$ (where $\lambda , \nu$ and $\mu$ are whole numbers), then find the value of $$\sum_\text{cyc} \lambda^\nu+\nu^\lambda$$

Simply opening up the expression yields that $G \cdot O \cdot E\cdot L=10^4$, Moreover, basic manipulations give: $$G=E=O/10$$ If we try to find the minimum value of the expression using the AM-GM Inequality then the powers of $2$ and $3$ won't come as whole numbers (which basically proves that somewhere I have faltered). Kindly help me out.

Michael Hardy
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Harsh Sharma
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1 Answers1

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I'm fairly sure you've done nothing wrong. As you realized, we can get $GOEL=10^4$ by adding the equations together. Now using AM-GM we have. $$\frac{3G+2L+2O+E}{4} \geq (12 \cdot GOEL)^\frac{1}{4}=(3^{\frac{1}{4}} \cdot 2^{3/2} \cdot 5^1)$$ I suspect that there is an error somewhere in either the question or your description of it, but the method that you used to solve is fine.

Isaac Browne
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