Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ 
Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ without using a calculator.

I rearraged to get $4 \cdot 1976^4-1 > 10^{-19.76} \cdot 16 \cdot 1976^7$ and so we have to prove that $4 \cdot 1976^4-1-10^{-19.76} \cdot 16 \cdot 1976^7>0$. How should I do that?

We have LHS is greater than $\frac{1}{5\cdot2000^3}-\frac{1}{10\cdot1000^7}>2.4\cdot10^{-11}$
I suspect the LHS should be the 3rd power instead of the 7th. The inequality is still true in that case, but a little more care is needed.
[For example you could approximate as $\frac{1}{4\cdot2000^3}-\frac{1}{10\cdot1500^3}=\frac{1}{32}10^{-9}-\frac{1}{3.375}10^{-10}>(\frac{1}{32}-\frac{1}{33})10^{-9}>10^{-13}$]
