While looking for trigonometric inequalities, I stumbled upon Aristarchus' inequality, which states that for $0<\alpha<\beta<\pi/2$


In this post (Proof of Aristarchus' Inequality) user141614 shows a completely algebraic proof of the first inequality using only $\sin(\alpha)<\alpha<\tan(\alpha)$.

I tried for a long time to reproduce a similar proof for $\tan$ by trying to prove the equivalent inequality $$\frac{\tan(\beta)-\tan(\alpha)}{\beta-\alpha}>\frac{\tan(\alpha)}{\alpha},$$ by using user141614's same idea, combined with trigonometric identities for the sum and product, but without success.

Does someone have hint on how to approach the problem? I really want an algebraic proof (which can potentially rely on easy-to-prove inequalities as the one above), no calculus.

Thank you in advance


I managed to find an easy proof:

For $0<\alpha<\beta<\pi/2$ we have $1>\cos(\alpha)>\cos(\beta)>0$ and thus (the orange chain was known) $$0<\sin(\alpha)\cos(\alpha)<\color{orange} {\sin(\alpha)<\alpha<\tan(\alpha)}=\frac{\sin(\alpha)}{\cos(\alpha)}<\color{red}{\frac{\sin(\alpha)}{\cos(\beta)}}.$$ Consequently, using that $0<\beta-\alpha<\beta$ in the previous equation (replacing $\alpha$ by $\beta-\alpha$) we obtain: $$\color{orange}{\beta-\alpha}<\color{red}{\frac{\sin(\beta-\alpha)}{\cos(\beta)}}=\tan(\beta)\cos(\alpha)-\sin(\alpha).$$ We conclude that \begin{align*} \frac{\beta}{\alpha}&=\frac{\beta-\alpha}{\alpha}+1\\ & <\frac{\tan(\beta)\cos(\alpha)-\sin(\alpha)}{\sin(\alpha)}+1\\ & = \frac{\tan(\beta)}{\tan(\alpha)}. \end{align*}


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