b)

Does someone have another solution for b and a solution for a without calculus? Perhaps by induction?

(a) & Telescope would solve (b).

Here my solution:a)If $a,b>0$,then binomial formula Newton gives $(a+b)^n>a^n+na^{n-1}b$ ,whence $(n+1)^n>2n^n$.The tenth degree of this inequality gives $(n+1)^{10n}>2^{10}n^{10n}>1000n^{10n}$ Logarithms, this inequality we obtain $\lg(n+1)>\frac{3}{10n}+{lgn}$ Ok _____________________________________ Azerbaijan Land of the Fire

Well for a) inequality is equivalent in showing $${(1 + \frac{1}{n})}^{10n} \geq 1000$$. But ${(1 + \frac{1}{n})}^ {n}$ is a decreasing function after $n \geq 3$ and reaches its limit $e$. So, we have $${({(1 + \frac{1}{n})}^{n})}^{10} > e^{10} > 2^{10} > 1000 $$.

Kunihiko_Chikaya wrote: (a) & Telescope would solve (b). Yes, you have right, just sumarize the a) and you get b). Or simple you can prove by induction!