Show $\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 5$ 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?
 A: $\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 4\times\sqrt[4]{\frac{\log 3}{\log 2}\times\frac{\log 4}{\log 3}\times\frac{\log 5}{\log 4}\times\frac{\log 6}{\log 5}}=4\times\sqrt[4]{\log_2 6}.$
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.
A: Converting to logs of a single base and using AGM, we have
$$\begin{align}
\log_23+\log_34+\log_45+\log_56
&={\log3\over\log2}+{\log4\over\log3}+{\log5\over\log4}+{\log6\over\log5}\\
&\ge4\sqrt[4]{{\log3\over\log2}\cdot{\log4\over\log3}\cdot{\log5\over\log4}\cdot{\log6\over\log5}}\\
&=4\sqrt[4]{\log6\over\log2}\\
&=4\sqrt[4]{1+{\log3\over\log2}}
\end{align}$$
So to show the desired inequality, it suffices to show that $256(1+\log3/\log2)\gt625$, or $\log3/\log2\gt369/256$. But
$$2\log3=\log9\gt\log8=3\log2$$
and
$$3\cdot256=768\gt738=2\cdot369$$
together tell us
$${\log3\over\log2}\gt{3\over2}\gt{369\over256}$$ as desired.
A: @Startwearingpurple: I did not downvote, but I thought your proposition is interesting because it implies the OP's result. Here is a proof, though it may be just useless.
Problem. Prove that $\log_23+\log_34+\log_45>4$.
Proof. Note that one has $$\log_23>1.58\Leftrightarrow 3^{50}>2^{79}\Leftrightarrow\left(\frac{3^8}{2^{13}}\right)^6>\frac 2 9,$$
$$\log_34>1.26\Leftrightarrow 4^{50}>3^{63}\Leftrightarrow\left(\frac{4^7}{3^{9}}\right)^7>\frac 1 4,$$ and
$$\log_45>1.16\Leftrightarrow 5^{25}>4^{29}\Leftrightarrow\left(\frac{5^5}{4^{6}}\right)^5>\frac 1 4.$$ It follows that $$\log_23+\log_34+\log_45>1.58+1.26+1.16=4.$$ QED
