# 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 :)

EDIT: $\log$ means $\log_{10}$

• If you want your solution to work, you need to show $$\displaystyle 5^{\log_{10}5} \gt 3 \text{ and } 7^{\log_{10}7} \gt 4$$ Oct 14, 2013 at 13:43
• I wonder whether $(\log_{10} x)^2 + (\log_{10} y)^2 > \log_{10}(x+y)$ always (or near $x=5$ and $y=7$).
– lhf
Oct 14, 2013 at 13:46
• Or you can show $\log_{10}(5) \gt \log_5(3)$ and $\log_{10}(7) \gt \log_7(4)$ Oct 14, 2013 at 13:51
• – lhf
Oct 14, 2013 at 13:55
• @Lhf $(\lg x)^2$ is a concave function when $x>1$. However, the statement still holds true for $x=1, y = 11$ (by computation), hence it holds true for all positive $x+y=12$. Oct 14, 2013 at 13:55

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 $3\cdot \log_{10}49> 5$.

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

$${\sf log}(5)^2+{\sf log}(7)^2 \geq \big(\frac{2}{3}\big)^2+ \big(\frac{5}{6}\big)^2=\frac{41}{36}=\frac{82}{72}\geq\frac{81}{72} \geq \frac{9}{8} \geq {\sf log}(12)$$