Given that $$left(1+frac{2}{n}right)^ngeqslantfrac{19}{3}-frac{6}{n}quadtext{for} quad ninmathbb{N},ngeqslant 3$$ and that $sqrt 3<frac{7}{4}$ 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 $sqrt 3<frac{7}{4}$, 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.

Thank you for your help.