<s>[INST] Prove the logarithmic inequality Prove that:
$(\log_{24}{48})^2+(\log_{12}{54})^2>4$
I tried to put $t=\log_23$ and get the equation $6t^4+32t^3+22t^2-84t-74>0$.
But I can't do anything with it...
 [/INST] Let $l=\log_{2}3$, so $\log_{3}2=\frac{1}{l}$ and $l<\frac{8}{5}$ since $2^8>3^5$, and $l>\frac{11}{7}$ since $2^{11}<3^7$.
Then $\displaystyle\log_{24}48=\frac{\log_{2}48}{\log_{2}24}=\frac{4+l}{3+l}>\frac{4+8/5}{3+8/5}=\frac{28}{23}\;\;\;$ and $\hspace{.36 in}\displaystyle\log_{12}54=\frac{\log_{3}54}{\log_{3}12}=\frac{3+\log_{3}2}{1+2\log_{3}2}=\frac{3+1/l}{1+2/l}=\frac{3l+1}{l+2}>\frac{3(\frac{11}{7})+1}{\frac{11}{7}+2}=\frac{8}{5}$, 
so $\displaystyle \left(\log_{24}48\right)^2+\left(\log_{12}54\right)^2>\left(\frac{28}{23}\right)^2+\left(\frac{8}{5}\right)^2>4$.
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