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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?

Aaron Hendrickson
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mathx
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    I'm not sure how you could use quadratic equations in this case, but you could prove this with the AM-GM inequality, and show that $\sqrt[4]{\log_2(6)}>\frac{5}{4}$. – projectilemotion Jun 02 '21 at 21:03
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    Similar: https://math.stackexchange.com/q/630489/42969 – Martin R Jun 02 '21 at 21:06
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    $$ \log_23 + \log_4 5 = \log_49+\log_45 = \log_445 > \log_432 = \frac 5 2 $$ – Michael Hardy Jun 02 '21 at 21:18
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    If one really insists on using quadratic equations, then maybe one could construct a similar argument as in https://artofproblemsolving.com/community/c4h2332013p18725522 (post #16)? In that post they prove the (rather strong) inequality $$\log_2 3+\log_3 4+\log_4 5>4.$$ – projectilemotion Jun 02 '21 at 21:44
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    @MichaelHardy, nice observation, but I'm not sure how much it helps, since $5/2+\log_34+\log_56\lt5$. – Barry Cipra Jun 02 '21 at 23:40

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

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.

Boshu
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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.

Barry Cipra
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@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

Pythagoras
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    I have no problem with downvotes. All such inequalities can be proved systematically by expanding the functions involved into Taylor series at suitably chosen points and using Lagrange remainder. Which is why one should simply use numerics whenever such problem really arises. I think constructing tailored solutions gives a wrong impression about what mathematics really is. – Start wearing purple Jun 03 '21 at 13:49