Prove that $$1-(4/3(\sin^3 \theta/2))/(\theta-\sin\theta)<(1-\cos\theta/2)(3/5-(3/1400)\pi^2/n^2)$$ holds for $0\le\theta\le\pi/2.$ Here $n$ is an integer greater than or equal to two.
This an unproved inequality due to Christian Huygens. I am preparing a paper on his approximations to $\pi$ and this is an inequality which he stated without proof in his famous work "Inventa." I have been unable to prove it and I would like to include a proof in my paper. Any help would be greatly appreciated.
$\newcommand\th\theta$This inequality is false for $n\ge2$. For instance, the difference between its left and right sides is $<-0.0000589\dots<0$ for $\theta=\pi/2$ and $n=2$ (and hence for all $n\ge2$).
We see that the inequality may be wrong "only slightly". Here is a correct version of the inequality. Replace $3/5-(3/1400)\pi^2/n^2$ in the OP by $c\in(0,3/5]$ and use the substitution $$t=\tan\frac\th4,\quad \sin\frac\th2=\frac{2t}{1+t^2}, \quad \cos\frac\th2=\frac{1-t^2}{1+t^2}, \quad \sin\th=\frac{4t(1-t^2)}{(1+t^2)^2}, \quad \th=4\tan^{-1}t$$ to rewrite the inequality in question as \begin{equation*} f(t):=f_c(t):=4 \tan ^{-1}(t) -\frac{4t \left(1-t^2\right)}{\left(1+t^2\right)^2} -\frac{32 t^3}{3 (1+t^2)^2 (1-(2 c-1) t^2)}>0 \tag{10}\label{10} \end{equation*} for all \begin{equation*} t\in(0,t_*],\quad t_*:=\tan\frac\pi8=\sqrt{2}-1. \end{equation*}
We have \begin{equation*} g(t):=g_c(t):=f'(t)\frac{3(1+t^2)^3(1-(2 c-1) t^2)^2}{64t^4} =3 - 5 c + (3 - 9 c + 6 c^2) t^2. \end{equation*} So, letting \begin{equation*} c_*:=\frac{1}{6} \left(12+5 \sqrt{2}-\sqrt{122+84 \sqrt{2}}\right)=0.59225\ldots, \end{equation*} we see that $f_c$ is increasing on $(0,t_*)$ if $c\in(0,c_*]$, and, for some $t_1\in(0,t_*)$, $f_c$ is increasing on $(0,t_1]$ and decreasing on $[t_1,t_*]$ if $c\in(c_*,3/5)$. Also, $f_c(0)=0$.
So, $f_c>0$ on $(0,t_*]$ iff $f_c(t_*)>0$, which latter occurs iff \begin{equation*} c<c_{**}:=\frac{12+4 \sqrt{2}-6 \pi }{12-6 \sqrt{2}-6 \pi +3 \sqrt{2} \pi } =0.59451\ldots. \end{equation*}
That is, the corrected inequality, with $c$ in place of $3/5-(3/1400)\pi^2/n^2$, holds for all $\th\in(0,\pi/2]$ iff $c<c_{**}=0.59451\ldots$; the latter number is just a bit less than $3/5$.
After this answer was posted, the OP changed the inequality to its opposite, thus invalidating the answer. The role of $n$ remains unexplained by the OP.
So, to answer the changed question, we will again replace $3/5-(3/1400)\pi^2/n^2$ in the OP by $c\in(0,3/5]$ and consider the inequality opposite to \eqref{10}:
\begin{equation*}
f(t)=f_c(t)<0 \tag{20}\label{20}
\end{equation*}
for all $t\in(0,t_*]$.
As noted above, $f_c(0)=0$ and $f_c$ is increasing in a right neighborhood of $0$ if $c\in(0,3/5)$. So, \eqref{20} fails to hold for all $t\in(0,t_*]$ if $c\in(0,3/5)$. On the other hand, \begin{equation} f'_{3/5}(t)=-\frac{128 t^6}{\left(5-t^2\right)^2 \left(t^2+1\right)^3}<0 \end{equation} for $t\in(0,t_*]$. Therefore and because $f_{3/5}(0)=0$, we see that \eqref{20} holds for $c=3/5$ and all $t\in(0,t_*]$ and, moreover, $3/5$ is the best (that is, the smallest) value of $c$ for which \eqref{20} holds for all $t\in(0,t_*]$.
Thus, inequality \eqref{10} holds for some $c\in(0,3/5]$ and all $t\in(0,t_*]$ iff $c<c_{**}=0.59451\ldots$, and the opposite inequality \eqref{20} holds for some $c\in(0,3/5]$ and all $t\in(0,t_*]$ iff $c=3/5=0.60000\ldots$. We see that the discrepancy between the optimal values $0.59451\ldots$ and $0.60000\ldots$ is small.
Prove that $$1-(4/3(\sin^3 \theta/2))/(\theta-\sin\theta)<(1-\cos\theta/2)(3/5-(3/1400)\pi^2/n^2)$$ holds for $0\le\theta\le\pi/2.$ Here $n$ is an integer greater than or equal to two.
I'm afraid the new inequality is also false. The function $f(\theta)$ = right-hand-side minus left-hand-side has the small-$\theta$ expansion $$f(\theta)=-\frac{3 \pi ^2\theta^2}{11200 n^2}+{\cal O}(\theta^4),$$ so it is negative at small $\theta$ for any integer $n$, violating the inequality.
I am curious to know what Huygens actually wrote, I am unable to extract the inequality from the Latin source. Perhaps he was merely pointing out a close approximation. (I note that the inequality is only violated slightly.)
After entering my addendum I noticed that Iosif Pinelis has reached the same conclusion in a separate post --- and moreover given a proof. All credit to him.
Addendum: Rescuing the Huygens identity
The appearance of the integer $n$ in Huygens' identity is puzzling, and the fact that it is mistaken is embarassing. I think the way to understand the identity is that $\theta$ and $n$ are not independent: Huygens writes his formulas with reference to an $n$-sided polygon inscribed in the unit circle, in terms of its perimeter $p_n=2n\sin\theta$, $\theta=\pi/n$. It therefore makes sense to understand the inequality upon substitution of $\pi/n\mapsto \theta$,
$$1-\frac{(4/3)\sin^3 \theta/2}{\theta-\sin\theta}<(1-\cos\theta/2)(3/5-(3/1400)\theta^2).$$ Now the small-$\theta$ expansion of the function $g(\theta)=$ right-hand-side minus left-hand-side is positive, $$g(\theta)=\frac{11}{2688000}\theta^6+{\cal O}(\theta^8).$$ Here is a plot of $g(\theta)$:
Defining the difference $$dif(\theta,n) =1-(4/3(\sin^3 \theta/2))/(\theta-\sin\theta)-(1-\cos\theta/2)(3/5-(3/1400)\pi^2/n^2)$$ We get approximately$$dif(1,2) = 0.000381$$ $$dif(\pi/2,2) = -0.000059$$ The first value is positive, so the conjectured inequality is false.
