Find the least value of $\log_a{bc}+\log_bac+\log_cab$

Question

If it is given $$\log_ac=\log_{\sqrt{a}}b+\log_ab^2+\log_{a^2}b^3+\log_{a^4}b^4+\log_{a^8}b^5+\cdots$$ and $a,b,c>1$ then find the least value of $$\log_a{bc}+\log_bac+\log_cab$$

Rewriting the first equation after doing some manipulation gives $$\log_ac=\log_{\sqrt{a}}b+\sum_{i=0}^{\infty}\log_{a^{2^i}}b^{i+2}$$ $$\implies \log_ac=2\log_{a}b+\log_ab\cdot\sum_{i=0}^{\infty}\frac{i+2}{2^i}$$ $$\implies \log_ac=2\log_{a}b+6\log_ab$$ $$\implies \log_ac=8\log_{a}b$$ $$\implies c=b^8$$ Now, $$\log_a{bc}+\log_bac+\log_cab=\log_ab+\log_ac+\log_ba+\log_bc+\log_ca+\log_cb$$ $$=\log_ab+\log_ac+\frac{1}{\log_ab}+\frac{1}{\log_ac}+8+\frac18$$ $$\ge2+2+8+\frac{1}{8}=\boxed{12.125}$$

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However, I don't think that it is the right answer as in the original question it was asked to find the value of $w+x+y+z$ where least value was equal to $\displaystyle\frac{w+x\sqrt{y}}{z}$ where $w,x,z$ are relatively prime and $y$ is not a square of any prime.

Am I right or wrong. Guide me if I'm wrong. Any help is greatly appreciated.

Answer 1

I think your general approach is correct, but the bound could have been a bit tighter:

When you have $\log_ab+\log_ac+\frac1{\log_ab}+\frac1{\log_ac}+8+\frac18$, you can use $c=b^8$ to obtain $$\log_ab+\log_ac+\frac1{\log_ab}+\frac1{\log_ac}+8+\frac18=\log_ab+8\log_ab+\frac1{\log_ab}+\frac1{8\log_ab}+8+\frac18$$$$=9\log_ab+\frac9{8\log_ab}+\frac{65}8.$$

Now $x+\frac1{8x}$ has minimum $\frac1{\sqrt2}$ so we obtain $$9\log_ab+\frac9{8\log_ab}+\frac{65}8\ge \frac9{\sqrt2}+\frac{65}8\gtrapprox14.48896$$

Answer 2

$$\log_{\sqrt a}b=\frac{\ln b}{(1/2)\ln a}=2\log_ab$$ $$\implies\log_ac=2\log_ab+\log_ab^2=4\log_ab\implies c=b^4$$ The expression being minimised then actually simplifies to $5\log_ab+\frac54\log_ba+\frac{17}4$, and so the minimum of $x+\frac1x$ being $2$ does not apply here. Let $x=\log_ab$, then the minimum of $x+\frac1{4x}$ is $1$ at $x=\frac12$, so the expression's minimum is $5+\frac{17}4=\frac{37}4$.