Show $\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 5$

Question

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?

Answer 1

$\log_2 3 + \log_3 4 + \log_4 5 + \log_5 6 > 4\times\sqrt[4]{\frac{\log 3}{\log 2}\times\frac{\log 4}{\log 3}\times\frac{\log 5}{\log 4}\times\frac{\log 6}{\log 5}}=4\times\sqrt[4]{\log_2 6}.$

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.

Answer 2

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.

Answer 3

@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