Compare $(5/6)^4$ and $(35/36)^{24}$ without calculation Can anyone provide some hint how to compare $(5/6)^4$ and $(35/36)^{24}$ without calculation? Thanks!
After some transformation this question is equivalent to check if $(5/6)^{20}(7/6)^{24}$ is larger than one or smaller than one.
 A: Raise both to the power of $3\over2$. Then we will be comparing
$\left(\frac{5}{6}\right)^6\;\text{ to }\;\left(\frac{35}{36}\right)^{36}$, and we might have heard a thing or two about the sequence $\left(\frac{n-1}{n}\right)^n$, which increases and converges to... well, that's not really important here, just note that it increases.
A: You are essentially asking whether 
$$\frac 56 \ \ \ \ \text{  or  }\ \ \ \ \left(\frac{35}{36}\right)^6$$
are bigger. 
Now by the Bernoulli's inequality,
$$\left(\frac{35}{36}\right)^6 = \left(1- \frac{1}{36}\right)^6 > 1 - 6\frac{1}{36} = \frac 56. $$
A: I think it would be more easy to find which is bigger by dividing one from another.
Let
\begin{align}
E&=\left(\frac{5}{6}\right)^4 \Big/\:{\left(\frac{35}{36}\right)^{24}}
\end{align}
so that
\begin{align}
E&=\left(5^4\!\cdot\!6^{48}\right)\big/\left(5^{24}\!\cdot\!7^{24}\!\cdot\!6^{4}\right)
=6^{44} \big/\left({7}^{24}\!\cdot\!{5}^{20}\right)<1
\end{align}
Since $\,E<1\,$ we have
\begin{align}
\left(\frac{5}{6}\right)^4 &\Big/{\left(\frac{35}{36}\right)^{24}}<1
&\implies&&
\bbox[1ex, border:2pt solid red]{\;\:
\left(\frac{5}{6}\right)^4 <{\left(\frac{35}{36}\right)^{24}}\;}
\end{align}
