2016 Eighth Romanian Master of Mathematics

If $x\ge 1-1/(100\cdot 2016)$, then $$x^{2016}\ge {\left(1-\frac{1}{100\cdot 2016}\right)}^{2016}>1-2016\cdot \frac{1}{100\cdot 2016}=1-\frac{1}{100}$$ by Bernoulli's inequality, whence the conclusion.

To establish the latter, refer again to Bernoulli's inequality to write $${\left(1+\frac{1}{99}\right)}^{2016}>{\left(1+\frac{1}{99}\right)}^{99\cdot 20}>{\left(1+99\cdot \frac{1}{99}\right)}^{20}=2^{20}>100\cdot 2016.$$