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?
 A: 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 $\frac{1}{4\cdot2000^3}-\frac{1}{10\cdot1500^3}=\frac{1}{32}10^{-9}-\frac{1}{3.375}10^{-10}>(\frac{1}{32}-\frac{1}{33})10^{-9}>10^{-13}$]
A: $$E=\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}-\frac{1}{10^{1444}}\gt 0\large?$$
$$E=1-\frac{1}{4\cdot 1976^4}-\frac{4\cdot 1976^3}{10^{1444}}\gt 0\large?$$
The answer is YES because $$\frac{1}{4\cdot 1976^3}\lt \frac 14$$ and because the numerator $4\cdot 1976^3$ has $11$ digits.
A: Seems really simple - am I missing something?
$a = 4\times 1976^3 < 4\times 2000^3 = 32\times 10^9$
$\implies 1/a > \frac{1}{32}10^{-9} > 3\times 10^{-11}$
$b= 16\times 1976^7 > 10^{22}$ and $1/b <10^{-22}$
$\implies \frac{1}{a} - \frac{1}{b} > 3 \times 10^{-11} - 10^{-22} >10^{-19.76}$  (howsoever you interpret that exponent)
