# algebra precalculus – Given some conditions, show that \$left(1+frac{2}{10}right)^{10}-3^{10/6}

Given that $$left(1+frac{2}{n}right)^ngeqslantfrac{19}{3}-frac{6}{n}quadtext{for} quad ninmathbb{N},ngeqslant 3$$ and that $$sqrt 3 and $$sqrt(6) 3 >frac{6}{5}$$, show that
$$left(1+frac{2}{10}right)^{10}-3^{10/6}<0.$$

My attempt:

Let $$n=10$$. It is then clear from the first given inequality that
$$left(1+frac{2}{10}right)^{10}geqslant frac{19}{3}-frac{6}{10}=frac{86}{15}.$$
Also,
$$-3^{10/6}=-3timessqrt{3}times sqrt(6) 3.$$
This is where I start to get stuck. Using $$sqrt 3, we can see that
$$-3timessqrt{3}times sqrt(6) 3>-frac{21}{4}times sqrt(6) 3$$ which doesn’t seem to help much. Alternatively, using $$sqrt(6) 3 >frac{6}{5}$$, we can see that
$$-3timessqrt{3}times sqrt(6) 3<-frac{18}{5}timessqrt 3$$
which also seems not to help. Even if we were to show that $$3^{10/6}$$ is greater than $$86/15$$ I don’t see how that would help as we wouldn’t know if it is smaller than
$$left(1+frac{2}{10}right)^{10}$$ or not.