Prove inequality $\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$

Question

I was trying to figure out if the following function can serve as a mean (see mean value theorem):

$$\arccos \left( \frac{\sin y-\sin x}{y-x} \right)$$

And turns out that for $x,y \leq \pi$ it does serve as a mean admirably.

But then I've noticed that for $0<x<1$ the following two functions are very close (see the picture):

Now how would you prove:

$$\arccos \left( \frac{\sin 1-\sin x}{1-x} \right) \leq \sqrt{\frac{1+x+x^2}{3}}$$

It's probably easier to consider another equivalent inequality:

$$\frac{\sin 1-\sin x}{1-x} \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$

Or even:

$$ \text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right) \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$

We could use Taylor series, but that's too cumbersome in my opinion.

Another way would be Mean value theorem itself, but I encounter the same problem.

Is there a simple way to prove this inequality?

My calculus is not as sharp as it used to be (just kidding, it was never sharp).

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Edit

Just to confirm (numerically) that the inequality holds, here is the plot of the difference between the two functions:

Answer 1

We have to show

$$\text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right) \geq (1-\frac{(1-x)^2}{24}) \cos \left(\frac{1+x}{2} \right) \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$

With the classic series for $\sin x$ and $0\leq x\leq 1$ it's clear that $$\text{sinc} \left(\frac{1-x}{2} \right) \geq 1-\frac{(1-x)^2}{24} $$

and therefore with a simple change of the first inequality it's now left

$$\frac{(1-x)^2}{24} \cos \left(\frac{1+x}{2} \right) \leq \cos \left(\frac{1+x}{2} \right)-\cos \sqrt{\frac{1+x+x^2}{3}}$$

or with $a:=\frac{1+x}{2}$ and $b:=\sqrt{\frac{1+x+x^2}{3}}$ and therefore $b\geq a$ it's

$$\frac{b^2-a^2}{2}\cos a \leq \cos a - \cos b$$

An equivalent inequality for this is $\int\limits_a^b (\sin x -x\cos a)dx \geq 0$.

$(A)\enspace$ Numerical proof with $0.5\leq a<b\leq 1$ for $\int\limits_a^b (\sin x -x\cos a)dx \geq 0$:

Increasing of $\sin x -x\cos a$: $(\sin x -x\cos a)'=\cos x - \cos a>0$ for $0\leq x<a$.

Decreasing with $a<x\leq b$.

Be $c:=\arccos(\sin(1))=0.570796…$ which means $\sin 1-1\cdot\cos c=0$.

(1) $\enspace c<a\leq 1$: $\enspace \sin x-x \cos a>0 \enspace$ for $\enspace a<x\leq b$

(2) $\enspace \frac{1}{2}\leq a\leq c$:

$\hspace{8mm}$ For every $a$ exists exactly one solution for $\sin x-x\cos a=0 \enspace$ when $\enspace a\leq x\leq b$ .

Definition: Be $x_0$ with $\sin x_0-x_0\cos \frac{1}{2}=0$

Because of $\enspace b=\sqrt{\frac{1-2a+4a^2}{3}}\enspace$ it’s $$\max\{b|\frac{1}{2}\leq a\leq c\}=\sqrt{\frac{1-2c+4c^2}{3}}=0.62226498459…<\frac{3}{4}<$$ $$<\min\{x\in[a;b]|\sin x-x \cos a=0 \text{ with }\frac{1}{2}\leq a\leq c\}=x_0=0.873…$$ => $\enspace \sin x-x \cos a>0$ for $a\leq x\leq b$

Therefore with (1)+(2) it’s $\int\limits_a^b (\sin x-x\cos a)dx \geq 0$ as expected.

In words: The integrand $\sin x-x\cos a$ is always positiv within the valid value area and therefore the integral too.

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$(B)\enspace$ A non-numerical proof using the first part of the explanations above:

We have to show that $\int\limits_a^b (\sin x-x\cos a)dx \geq 0$ .

This is true if $\enspace\sin x-x\cos a\geq 0\enspace$ for $\enspace a\leq x\leq b$ .

$\sin x-x\cos a\enspace$ is decreasing for $\enspace a<x<b\enspace$ because of $\enspace\displaystyle \frac{d}{dx}(\sin x-x\cos a)<0\enspace$ and therefore is

$\hspace{1cm}$ $\min\{\sin x-x\cos a|a\le x\le b\}=\sin b-b\cos a$

$\hspace{1cm}$ for $\enspace 0.5\leq a\leq b\leq 1\enspace$ with $\enspace b=\sqrt{\frac{1-2a+4a^2}{3}}$ .

$=>\enspace$ It has to be shown that $\enspace\displaystyle \cos a<\frac{\sin b}{b}\enspace$ e.g. by proving $\enspace\displaystyle\cos a<1.1-0.4 a<\frac{\sin b}{b}$ .

The left side is clear for $\enspace\displaystyle\frac{1}{2}\leq a\leq 1\enspace$ and the right side can be better handled if $\enspace a\enspace$ is substituted by $\enspace\displaystyle\frac{1}{4}(1+\sqrt{3}\sqrt{(2b)^2-1})\enspace$ with $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq b\leq 1\enspace$ so that we can simplify e.g. $\enspace\displaystyle 1-0.1\sqrt{3}\sqrt{(2b)^2-1}<1.15-0.4 b<\frac{\sin b}{b}$ .

It's $\enspace\displaystyle 1-0.1\sqrt{3}\sqrt{(2x)^2-1}<1.15-0.4 x\enspace$ true for $\enspace\displaystyle |x-1.5|<\frac{1}{4}\sqrt{15}\enspace$ which includes $\enspace\displaystyle \frac{1}{\sqrt{3}}\leq x\leq 1$ .

To verify $\enspace\displaystyle \cos a<\frac{\sin b}{b}\enspace$ we can use the classical series of $\enspace\cos\enspace$ and $\enspace\sin\enspace$ and get the following inequalities which have to be proved:

$\hspace{8mm}\displaystyle\cos x<1-\frac{x^2}{2}+\frac{x^4}{24}<1.1-0.4 x\enspace$ for $\enspace\displaystyle\frac{1}{2}\leq x\leq 1\enspace$ and

$\hspace{8mm}\displaystyle 1.15-0.4 x<1-\frac{x^2}{6}<\frac{\sin x}{x}\enspace$ for $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1\enspace$

$(1)\enspace\displaystyle \cos x<1-\frac{x^2}{2}+\frac{x^4}{24}\enspace$ for $\enspace\displaystyle\frac{1}{2}\leq x\leq 1$ :

$\hspace{8mm}$ This is true with $\enspace\displaystyle\cos x=\sum\limits_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}$

$\hspace{8mm}$ because of $\enspace\displaystyle\frac{x^{2k}}{(2k)!}-\frac{x^{2k+2}}{(2k+2)!}>0\enspace$ for $\enspace k\in\mathbb{N}$ .

$(2)\enspace\displaystyle 1-\frac{x^2}{2}+\frac{x^4}{24}<1.1-0.4 x\enspace$ for $\enspace\displaystyle\frac{1}{2}\leq x\leq 1$ :

$\hspace{8mm}$ This is true for $\enspace\displaystyle x:=\min{x}=\frac{1}{2}$

$\hspace{8mm}$ and because of $\enspace\displaystyle\frac{d}{dx}(1-\frac{x^2}{2}+\frac{x^4}{24})<\frac{d}{dx}(1.1-0.4 x)<0$ .

$(3)\enspace\displaystyle 1.15-0.4 x<1-\frac{x^2}{6}\enspace$ for $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1\enspace$ :

$\hspace{8mm}$ This is true for $\enspace\displaystyle x:=\min{x}=\frac{1}{\sqrt{3}}$

$\hspace{8mm}$ and because of $\enspace\displaystyle\frac{d}{dx}(1.15-0.4 x)<\frac{d}{dx}(1-\frac{x^2}{6})<0$ .

$(4)\enspace\displaystyle 1-\frac{x^2}{6}<\frac{\sin x}{x}\enspace$ for $\enspace\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1\enspace$ :

$\hspace{8mm}$ This is true with $\enspace\displaystyle \frac{\sin x}{x} =\sum\limits_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k+1)!}$

$\hspace{8mm}$ because of $\enspace\displaystyle\frac{x^{2k}}{(2k+1)!}-\frac{x^{2k+2}}{(2k+3)!}>0\enspace$ for $\enspace k\in\mathbb{N}$ .

With the verification of $(1)$ to $(4)$ the proof is completed.

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A summary of the steps of $(B)$ .

$\displaystyle sinc(\frac{1-x}{2})\cos(\frac{1+x}{2})\geq \cos\sqrt{\frac{1+x+x^2}{3}}$ is verified by proofs for $(1)$ and $(2)$ .

$(1)\enspace$ $\displaystyle sinc(\frac{1-x}{2})\geq 1-\frac{(1-x)^2}{24}\enspace$ (proof with series expansion)

$(2)\enspace$ $\displaystyle (1-\frac{(1-x)^2}{24})\cos(\frac{1+x}{2})\geq\cos\sqrt{\frac{1+x+x^2}{3}}\enspace$ (verified by the proof for $(3)$)

With $\enspace\displaystyle a:=\frac{1+x}{2}\in [\frac{1}{2};1]\enspace$ and $\enspace\displaystyle b:=\sqrt{\frac{1+x+x^2}{3}}\in [\frac{1}{\sqrt{3}};1]\enspace$ point $(2)$ changes to

$(3)\enspace$ $\displaystyle\int\limits_a^b (\sin x-x\cos a)dx\geq 0$ .

Because of $\enspace\min(\sin x-x\cos a)|_{a\leq x\leq b}=\sin b-b\cos a\enspace$ (proof by derivation) point $(3)$

is verified by the proof for $\enspace\displaystyle\cos a<\frac{\sin b}{b}\enspace$, points $(4)$ till $(8)$ .

$\displaystyle\frac{1}{2}\leq x\leq 1$ :

$(4)\enspace$ $\displaystyle \cos x<1-\frac{x^2}{2}+\frac{x^4}{24}\enspace$ (proof with series expansion)

$(5)\enspace$ $\displaystyle 1-\frac{x^2}{2}+\frac{x^4}{24}<1.1-0.4x\enspace$ (proof with derivation)

$\displaystyle\frac{1}{\sqrt{3}}\leq x\leq 1$ :

$(6)\enspace$ $\displaystyle 1-0.1\sqrt{3}\sqrt{(2x)^2-1}<1.15-0.4 x\enspace$ (proof with solving the quadratic equation)

$(7)\enspace$ $\displaystyle 1.15-0.4 x<1-\frac{x^2}{6}\enspace$ (proof with derivation)

$(8)\enspace$ $\displaystyle 1-\frac{x^2}{6}<\frac{\sin x}{x}\enspace$ (proof with series expansion)

Note: $(6)$ and $(7)$ can be put together; I haven't, for a better overfiew.

Answer 2

Assuming radians: This is a good question. Using $\sin a - \sin b = 2\sin \frac{a-b}2 \cos \frac{a+b}2$ gives:

$$ \frac 2{1-x} \sin \frac{1-x}2 \cos \frac{1+x}2 \le \cos \sqrt \frac {(1+x)^2-x}3 $$

Then using (1+x)/2=u:

$$ \frac 1u \sin (1-u) \cos u \le \cos \sqrt \frac {4u^2-2u+1}3$$

Then there is an identity you could use on the left side that I can't remember...

You could also prove that: $$ \sin 1 \le (1-x) \cos \sqrt \frac {1+x+x^2}3 +\sin x $$

By differentiating the right side, finding minima and showing that they are all >sin1, but this is equally as hard (if not harder) as expanding everything as series.

Answer 3

This is not a full answer, just a possible way to prove the inequality.

We use the following form of the inequality:

$$\text{sinc} \left(\frac{1-x}{2} \right) \cos \left(\frac{1+x}{2} \right) \geq \cos \sqrt{\frac{1+x+x^2}{3}}$$

It makes sense to try infinite products, mainly because we get rid of the square root:

$$\text{sinc}(t)=\prod_{n=1}^\infty \left(1-\frac{t^2}{\pi^2 n^2} \right)$$

$$\cos (t)=\prod_{n=1}^\infty \left(1-\frac{t^2}{\pi^2 (n-1/2)^2} \right)$$

Thus, our inequality becomes:

$$\prod_{n=1}^\infty \left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq \prod_{n=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)$$

Note that for $x<1$ every term in the infinite products is positive.

If, for example, every term of the product on the left is greater than every term of the product on the right, then our inequality is proven.

$$\left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq^? 1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2}$$

After expanding and simplifying I obtained the following:

$$-\left(4 \pi^2 n(2n-3)+3(\pi^2-(1+x)^2) \right)(1-x)^2 \geq^? 0$$

And this is highly questionable, i.e. not correct for most cases. Remember, we are interested in the case $x < 1$, so:

$$4 \pi^2 n(2n-3)+3(\pi^2-(1+x)^2) \leq^? 0$$

$$4 \pi^2 n(2n-3)+3\pi^2 \leq^? 3 (1+x)^2$$

For $n=1$ we have a trivial inequality:

$$-\pi^2 < 3 (1+x)^2$$

But for $n \geq 2$ the inequality quickly stops working.

So this method probably doesn't prove anything.

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On the other hand, we can compare the two term products on each side, i.e. prove that:

$$\prod_{n=k}^{k+1} \left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) \geq \prod_{n=k}^{k+1} \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)$$

If that fails, we can try $3$ product terms and so on. This is just some algebra that a CAS can take care of even for a large number of terms.

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Update

I decided to rearrange the second product so the terms are of the same order in $x$ and $n$:

$$\prod_{n=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (n-1/2)^2} \right)=\prod_{k=1}^\infty \left(1-\frac{1+x+x^2}{3\pi^2 (2k-3/2)^2} \right) \left(1-\frac{1+x+x^2}{3\pi^2 (2k-1/2)^2} \right)$$

Still, we have the same relation: only the first terms of the products obey the inequality, while all the rest seem to break it.

Below you can see the plot for:

$$ \dfrac{\left(1-\frac{(1-x)^2}{4\pi^2 n^2} \right) \left(1-\frac{(1+x)^2}{4\pi^2 (n-1/2)^2} \right) }{ \left(1-\frac{1+x+x^2}{3\pi^2 (2n-3/2)^2} \right) \left(1-\frac{1+x+x^2}{3\pi^2 (2n-1/2)^2} \right) }$$

For $x \in [0,1]$ and $n=1,2,3, \dots$.

I see no way to prove that the product of all terms for $n \geq 2$ is still closer to $1$ than the first term (despite numerical evidence), so this way doesn't seem to work as a proof of the original inequality.

Answer 4

To prove the inequality, it is enough to show that the inequality becomes equality at a single point (namely $x=1$).

To see this, proceed by contradiction. Suppose that the inequality does not hold at some point $x\in[0,1]$. Then, since all functions considered are continuous, it follows it doesn't hold on an interval. On the other hand, by a straightforward calculation, there exist points $y<1<z$ such that the inequality holds at $x=y$ and $x=z$ with strict inequality. Therefore, by the Intermediate Value Theorem there must be at least two points that make the inequality an equality.

In this direction, it might be useful to consider the inequality as an optimization problem, and prove that it can only have one optimizer (i.e. one that makes the inequality an equality). Both $\sin$ and $\cos$ are concave functions on $[0,1]$, and $\sqrt x$ is also concave on $[0,1]$. I imagine one can put these facts together into a solution...

Answer 5

Geometric Explanation “Why $3$ ?”

(Not an Answer)

In order to explain why the inequality could be true, we have: $$ x^3-1=(x-1)(1+x+x^2) \Rightarrow \cos\sqrt{\frac{1+x+x^2}{3}}=\cos\sqrt{\frac{1}{\color{red}{3}}\frac{x^{\color{red}{3}}-1}{x-1}} $$ And by considering the general case $\space x^n-1=(x-1)(1+x+x^2+\cdots+x^{n-1})$, Let: $$ \begin{align} & \color{red}{f_{\alpha}(x)} = \frac{\sin(1)-\sin(x)}{1-x}-\cos\sqrt{\frac{1}{\color{red}{\alpha}}\frac{x^{\color{red}{\alpha}}-1}{x-1}} \quad\colon\space \alpha \ge 1, \space \alpha \in \mathbb{R} \\[2mm] & \qquad \Rightarrow \space f_{\alpha}(0) = \lim_{x\rightarrow0}f_{\alpha}(x) = \sin(1)-\cos\left(1/\sqrt{\alpha}\right) \\ & \qquad \rightarrow \space \text{for}\space f_{\alpha}(0)=0 \Rightarrow \alpha=1/\arccos^2\left(\sin(1)\right) \\[2mm] & \qquad \space\&\space\space \space f_{\alpha}(1)= \lim_{x\rightarrow1}f_{\alpha}(x) = 0 \quad \left\{\text{for all}\space\alpha\right\} \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align} $$ Also: $$ \begin{align} & \color{red}{f'_{\alpha}(x)} = \small \frac{\sin(1)-\sin(x)-(1-x)\cos(x)}{(1-x)^2}-\frac{1-x^\alpha-\alpha(1-x)x^{\alpha-1}}{2\alpha(1-x)^2}\sqrt{{\alpha}\frac{x-1}{x^\alpha-1}}\,\sin\sqrt{\frac{1}{\alpha}\frac{x^\alpha-1}{x-1}} \\[2mm] & \qquad \Rightarrow \space f'_{\alpha}(0) = \lim_{x\rightarrow0}f'_{\alpha}(x) = \sin(1)-1+\frac{1}{\sqrt{\alpha}}\,\sin\left(1/\sqrt{\alpha}\right) \\[2mm] & \qquad \space\&\space\space \space f'_{\alpha}(1)= \lim_{x\rightarrow1}f'_{\alpha}(x) = \color{red}{\frac{\sin(1)}{4}\,(\alpha-3)} \\ & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align} $$

By swiping $\alpha$ from $1$ towards $\infty$, considering the range $x\in[0,1]$:

${\bf1}$. Starting by $\space f_1(x)=\left[\sin(1)-\sin(x)\right]/[1-x]-\cos(1)$, a well defined function,
$\qquad$ smooth decreasing curve from $(0,\sin1-\cos1)$ to $(1,0)$, no $x$ axis intersection (no zeros).

${\bf2}$. Smooth adjustment in the decreasing behavior while $\alpha$ increases.

${\bf3}$. At a certain value $1\lt\alpha_0\lt1/\arccos^2\left(\sin(1)\right)$, $f_{\alpha}(x)$ will stop being completely decreasing,
$\qquad$ start to have an increasing part, causing intersection(s) with $x$ axis (at least one zero).

${\bf4}$. Between $\alpha_0\lt\alpha\lt1/\arccos^2\left(\sin(1)\right)$, $f_{\alpha}(x)$ surly has at least one zero.

${\bf5}$. For $\alpha\gt1/\arccos^2\left(\sin(1)\right)\Rightarrow f_{\alpha}(0)\lt0$,
$\qquad$ and the function should stop intersecting with $x$ axis.

${\bf6}$. When $\alpha\rightarrow\infty\space\Rightarrow(x^\alpha-1)/(x-1)\approx1/(1-x) \Rightarrow f_{\alpha}(x)\approx1\frac{\sin(1)-\sin(x)}{1-x}-\cos\left(\frac{1/\alpha}{1-x}\right)$.


And from the definition of $f'_{\alpha}(x)$, we can calculate the value of $\alpha_0$ that stops the completely decreasing behavior and create a horizontal tangent at the end of the interval $x\in[0,1]$: $$ f'_{\alpha_0}(1)=0 \space\Rightarrow\space \frac{\sin(1)}{4}\left(\alpha_0-3\right)=0 \space\Rightarrow\space \color{red}{\alpha_0=3} $$ The inequality is really interesting because it shows the maximum $\alpha$ of $\left\{f_{\alpha}(x)\ge0\colon x\in[0,1]\right\}$
As-well-as, for $\left\{3\lt\alpha\lt1/\arccos^2\left(\sin(1)\right)\right\}$ we have $\left\{f_{\alpha}(0)>0 \space\&\space f'_{\alpha}(1)>0\right\}$, Thus:
The function $f_{\alpha}(x)$ have an odd number of zeros in the range $x\in[0,1]$.
In fact, if we can argue to replace the statement "at least one zero" with "at most one zero" for $3\lt\alpha\lt1/\arccos^2\left(\sin(1)\right)$, then the equality is proved!

Answer 6

Supplement to user90369's very nice answer.

It suffices to prove that, for all $x \in (0, 1)$, $$\frac{2\sin \frac{1 - x}{2}\, \cos \frac{1 + x}{2}}{1 - x} \ge \cos \sqrt{\frac{1 + x + x^2}{3}}. \tag{1}$$

Using the known inequality $\sin y \ge y - \frac{y^3}{6}$ for all $y \ge 0$, letting $y = \frac{1 - x}{2}$, we have \begin{align*} \sin \frac{1 - x}{2}\ge \frac{1 - x}{2} - \frac{1}{6}\left(\frac{1 - x}{2}\right)^3. \tag{2} \end{align*}

From (1) and (2), it suffices to prove that $$\left(1 - \frac{1}{24}(1 - x)^2\right) \cos \frac{1 + x}{2} \ge \cos \sqrt{\frac{1 + x + x^2}{3}}. \tag{3}$$

Letting $a := \frac{1 + x}{2}$ and $b := \sqrt{\frac{1 + x + x^2}{3}}$. We have $b^2 - a^2 = \frac{1}{12}(1 - x)^2$. (3) is written as $$\left(1 - \frac{b^2 - a^2}{2}\right)\cos a \ge \cos b$$ or $$2\sin \frac{b + a}{2} \, \sin \frac{b - a}{2} \ge \frac{b^2 - a^2}{2}\cos a. \tag{4}$$ Note that $1/2 \le a < b \le 1$.

Using $\sin u \ge (4\sin \frac14) u$ for all $u \in [0, 1/4]$ (easy to prove), letting $u = \frac{b - a}{2} \in [0, 1/4]$, we have $$\sin \frac{b - a}{2} \ge (4 \sin \frac14)\frac{b - a}{2}. \tag{5}$$

From (4) and (5), it suffices to prove that $$2\sin \frac{b + a}{2}\cdot 4 \sin \frac14 \ge (b + a)\cos a. $$

Let $f(b) := 2\sin \frac{b + a}{2}\cdot 4 \sin \frac14 - (b + a)\cos a$. We have $$f'(b) = \cos \frac{b + a}{2} \cdot 4\sin \frac14 - \cos a \le \cos \frac{a + a}{2} \cdot 4\sin \frac14 - \cos a < 0.$$ Also, we have, for all $1/2 \le a \le 1$, \begin{align*} f(1) &= 2\sin \frac{1 + a}{2}\cdot 4 \sin \frac14 - (1 + a)\cos a\\ &\ge 2\sin \frac{1 + \frac12}{2}\cdot 4 \sin \frac14 - (1 + a)\left(\cos \frac12 - (a - 1/2)\sin \frac12\right)\\ &= \sin \frac12 \cdot a^2 + \left(\frac12 \sin \frac12 - \cos\frac12\right)a + 8\sin \frac34 \, \sin \frac14 - \cos \frac12 - \frac12 \sin \frac12\\ &> 0 \end{align*} where we use $\cos a \le \cos \frac12 - (a - 1/2)\sin \frac12$ since $x\mapsto \cos x$ is concave on $[0, 1]$. Thus, $f(b) \ge 0$ for all $1/2 \le a < b \le 1$.

We are done.