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 $\log_{10}$
 A: 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 $3\cdot \log_{10}49> 5$.
A: From $5^3=125$ and $7^6=117649$, we deduce that ${\sf log}(5) \geq \frac{2}{3}$
and ${\sf log}(7) \geq \frac{5}{6}$.
From $3(6^7)=839808$ and $5^9=1953125$, we deduce that $3(6^7) \leq 5^9$ and hence
$12^8 \leq 10^9$. So  ${\sf log}(12) \leq \frac{9}{8}$. 
Finally, we have
$$
{\sf log}(5)^2+{\sf log}(7)^2 \geq \big(\frac{2}{3}\big)^2+
\big(\frac{5}{6}\big)^2=\frac{41}{36}=\frac{82}{72}\geq\frac{81}{72} \geq \frac{9}{8} \geq {\sf log}(12)
$$
