How would you show
$$\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 5?$$
After trying to represent the mentioned expression in the same base, a messy expression is created.
The hint mentioned that the proof can take help of quadratic equations.
Could you provide some input?
Is it a good idea to think of some graphical solution?
Best Answer
$\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 4\times\sqrt[4]{\frac{\log 3}{\log 2}\times\frac{\log 4}{\log 3}\times\frac{\log 5}{\log 4}\times\frac{\log 6}{\log 5}}=4\times\sqrt[4]{\log_2 6}.$
Now, $5/4=1.25$. So we are done if we can show that $1.25^4<2.45<\log_2 6$. To see this, observe that $2^{2.5}=\sqrt{2^5}=\sqrt 32<6$. That gives you the result.