Which is Greater: $\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?

[anemone]

State with reason which of these is greater?

$\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?

 

[anemone]

State with reason which of these is greater?

$\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?

Hint:

Spoiler

Extended Cauchy-Schwarz Inequality

 

[topsquark]

My calculator was faster. (Yes)

-Dan

 

[anemone]

State with reason which of these is greater?

$\log_{10}12$ or $(\log_{10}5)^2+(\log_{10}7)^2$?

My solution:

Spoiler

Note that by using the extended Cauchy-Schwarz inequality, we have

$\dfrac{(\log_{10}5)^2}{1}+\dfrac{(\log_{10}7)^2}{1}\ge \dfrac{(\log_{10}5+\log_{10}7)^2}{1+1}=\dfrac{(\log_{10}35)^2}{2}$

But we know $35^2=1225>10^3$, so $\log_{10}35>\dfrac{3}{2}$ and therefore $\dfrac{(\log_{10}35)^2}{2}>\dfrac{9}{8}$.

If we can prove $\dfrac{9}{8}>\log_{10} 12$, then we can conclude $(\log_{10}5)^2+(\log_{10}7)^2>\log_{10}12$.

Observe that $10^3=1000>864=\dfrac{12^3}{2}$, so we get $10^9>\dfrac{12^9}{8}$, since $\dfrac{12^9}{8}>12^8$, we have proved $10^9>12^8$, i.e. $\dfrac{9}{8}>\log_{10} 12$.

Therefore, $\dfrac{(\log_{10}5)^2}{1}+\dfrac{(\log_{10}7)^2}{1}\ge\dfrac{(\log_{10}35)^2}{2}\ge\dfrac{9}{8}>\log_{10} 12$.