<s>[INST] Prove that $\log^25 + \log^27 > \log12$. Prove that $\log^25 + \log^27 > \log12$.
What I tried so far:
$\log^25 + \log^27 > \log3 + \log4$
$(\log5 + \log7)^2 - 2 \cdot \log5 \cdot\log7 > \log3 + \log4$
But it seems that I'm not even near the result.
Every suggestion / hint would be appreciated :)
Thanks in advance.
EDIT: $\log$ means $\log_{10}$
 [/INST] Without actually computing exact logs,...
 $$\log_{10}5 \cdot \log_{10}5 + \log_{10}7\cdot \log_{10}7 > \log_{10}12$$
$$\iff \frac{\log_{10}5}{\log_{10}7} + \frac{\log_{10}7}{\log_{10}5} >  \frac{\log_{10}12}{\log_{10}5 \cdot \log_{10}7}$$
Now LHS $>2$ as it is the sum of a positive number ($\neq 1$) and its reciprocal.  So it is sufficient to show that RHS $< 2$, which is equivalent to:
$$\log_{10}12 < 2\log_{10}5 \cdot \log_{10}7 \iff \log_{5}12 < \log_{10}49 \iff 3\cdot\log_{5}12 < 3\cdot \log_{10}49$$
But $12^3 = 1728 < 5^5$, while $49^3 > 10^5$ shows $3\cdot\log_{5}12< 5$, while $3\cdot \log_{10}49> 5$.
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