# Prove that $16(a\sin a + \cos a - 1)^2 \le 2a^4 + a^3 \sin 2a, \ \forall a\ge 0$

Problem 1: Prove that $$16(a\sin a + \cos a - 1)^2 \le 2a^4 + a^3 \sin 2a, \ \forall a\ge 0.\tag{1}$$

This is the stronger version of the following Prove that $12(a\sin a+\cos a-1)^2\le 2a^4+a^3\sin(2a)$,$\forall a\in (0,\infty)$:

Problem 2: Prove that $$12(a\sin a + \cos a - 1)^2 \le 2a^4 + a^3 \sin 2a, \ \forall a\ge 0. \tag{2}$$

For Problem 2, there is a very nice solution using Cauchy-Bunyakovsky-Schwarz inequality for integral Prove that $12(a\sin a+\cos a-1)^2\le 2a^4+a^3\sin(2a)$,$\forall a\in (0,\infty)$. Indeed, the inequality (2) is nothing but $$12\left(\int_0^a x\cos x \mathrm{d} x \right)^2 \le 12\left(\int_0^a x^2 \mathrm{d} x\right) \left(\int_0^a \cos^2 x \mathrm{d}x\right).$$

For Problem 2, I gave a complicated proof. See Prove that $12(a\sin a+\cos a-1)^2\le 2a^4+a^3\sin(2a)$,$\forall a\in (0,\infty)$

Are there any nice solutions for Problem 1?

Any comments and solutions are welcome and appreciated.

• How did you obtain the value 16? Commented Jan 30, 2021 at 18:28
• @N.S.JOHN Thanks for comment. The first positive zero of $a\sin a + \cos a - 1$ is $a_1 \approx 1.33$. Graphically, $\frac{2a^4+a^3\sin(2a)}{(a\sin(a)+\cos(a)-1)^2}$ is strictly increasing on $(0, a_1)$. Also, $\lim_{a\to 0}\frac{2a^4+a^3\sin(2a)}{(a\sin(a)+\cos(a)-1)^2} = 16$. Second, graphically, $f(a) = 2a^4 + a^3 \sin 2a - 16(a\sin a + \cos a - 1)^2$ is strictly increasing on $a > 0$. $f'(a) = 2a((2\cos^2 a + 3)a^2 - 13a\sin a \cos a - 16\cos^2 a + 16\cos a)$. We need to prove $(2\cos^2 a + 3)a^2 - 13a\sin a \cos a - 16\cos^2 a + 16\cos a \ge 0$. We may use my approach in the link. Commented Jan 31, 2021 at 1:07
• In the nice solution to problem 2, the application of Cauchy-Schwarz is far from tight, since $x$ and $\cos x$ are not proportional. Maybe you can use a Cauchy-Schwarz defect inequality. Commented Feb 10, 2021 at 9:21
• @mathworker21 Thanks for valuable comment. Commented Feb 10, 2021 at 11:12
• The command of Mathematica 12.2 Minimize[{2 a^4 + a^3*Sin[2 a] - 16*(a*Sin[a] + Cos[a] - 1)^2, a >= 0}, a] produces {0, {a -> 0}}. Commented Feb 13, 2021 at 8:40

$$\color{brown}{\textbf{Preliminary notes.}}$$

Dividing of the given inequality $$(1)$$ to $$\;4a^4\;$$ allows to present it in the equivalent form of $$\left(2\text{ sinc } a - \text{ sinc}^2\,\dfrac a2\right)^2 \le \dfrac{1+\text{sinc }2a}2,\tag1$$ or $$\;f_1(a)\le f_2(a),\;$$ where $$f_1(x) = \left(2\text{ sinc } x - \text{ sinc}^2\,\dfrac x2\right)^2,\quad f_2(x)\le \dfrac{1+\text{sinc }2x}2.\tag2$$

It suffices to prove the inequalities $$f_1(x) \le (1-\,^1\!/_6\,x^2)^2\le f_2(x),\quad \text{if}\quad x\in\left[0,\dfrac\pi2\right],\tag3$$ and $$f_1(x)\le \dfrac{4\pi-3}{8\pi}\le f_2(x),\quad \text{if}\quad x\in\left(\dfrac\pi2,\infty\right).\tag4$$

We can obtain the derivative $$f_1'(x)= -\dfrac{8}{x^5}\,\sin x\left(\tan \dfrac x2-x\right)\bigg((x^2-2) \cos x - 2 x \sin x+2\bigg)\tag5$$

The root of the derivative ar $$\;x=r\approx\dfrac73\;$$ corresponds to the root of $$\;f_1.$$

The root of the derivative near $$\;x=m\approx 4\;$$ corresponds to the secondary maximum of $$\;f_1.$$

$$\color{brown}{\mathbf{\text{Interval }\left[0,\dfrac\pi2\right].}}$$

Taking in account inequality $$\;x^2<3,\;$$ one can get $$2\text{ sinc } x-\text{ sinc }^2\,\dfrac x2 = 2\left(1-\dfrac16x^2+\dfrac1{5!}x^4-\dfrac1{7!}x^6+\dots\right) - \left(1-\dfrac16\left(\dfrac x2\right)^2+\dfrac1{5!}\left(\dfrac x2\right)^4+\dots\right)^2$$$$\le 2-\dfrac13x^2+\dfrac{x^4}{60} - \left(1-\dfrac{x^2}{24}\right)^2 \le 1-\dfrac14x^2+\dfrac1{36}x^4 \le 1-\dfrac14x^2+\dfrac1{12}x^2= 1-\dfrac{x^2}6,$$

$$f_1(x)\le(1-\,^1\!/_6 x^2)^2.$$

Om the other hand, $$\dfrac12\big(1+\text{ sinc } 2x\big)- (1-\,^1\!/_6 x^2)^2 = 1-\dfrac1{12}\,(2x)^2+\dfrac1{2\cdot5!}(2x)^4-\dfrac1{2\cdot7!}(2x)^6+\dots -1 + \dfrac13x^2 - \dfrac1{36}x^4$$$$\ge\left(\dfrac1{15}-\dfrac1{36}\right)x^4\ge 0,$$$$f_2(x)\ge (1+\,^1\!/_6x^2)^2.$$ Therefore, inequality $$(3)$$ is proved.

$$\color{brown}{\mathbf{\text{Interval }\left(\dfrac\pi2,\infty\right).}}$$

Since $$f_1\left(\dfrac\pi2\right)=\left(\dfrac4\pi-\dfrac8{\pi^2}\right)^2 <\dfrac14<\dfrac{4\pi-3}{8\pi},$$ then $$f_1(x)\le \dfrac{4\pi-3}{8\pi},\quad \text{if}\quad x\in\left(\dfrac\pi2,r\right].$$

At the same time, $$f'_1(4.0)>0,\;f'_1(4.1)<0\quad\Rightarrow\quad m\in(4.0,4.1).$$ Then $$f_1(m) - \dfrac{4\pi-3}{8\pi} < (\text{ sinc}^2 2 - 2\text{ sinc }4.1)^2 - \dfrac{4\pi-3}{8\pi} < 0,$$ and $$f_1(x)\le \dfrac{4\pi-3}{8\pi},\quad \text{if}\quad x\in\left(\dfrac\pi2,\infty\right].\tag6$$

The least value of $$\;\text{ sinc } x\;$$ is situated near $$\;x=\dfrac{3\pi}2,\;$$ wherein $$\text{ sinc }\dfrac{3\pi}2 = -\dfrac2{3\pi},\quad \text{ sinc }\dfrac{4\pi}3 = -\dfrac{3\sqrt3}{8\pi}>\text{ sinc }\dfrac{3\pi}2,\quad \text{ sinc }\dfrac{5\pi}3 = -\dfrac{3\sqrt3}{10\pi}>\text{ sinc }\dfrac{3\pi}2.$$ Taking in account unimodality of $$\;\text{ sinc }x\;$$ in$$\;\left[\dfrac{4\pi}3,\dfrac{5\pi}3\right],\;$$ the least value of $$\;\text{ sinc }x\;$$ belongs to this interval. Therefore, $$\text{ sinc }x>-\dfrac{\sin\dfrac{3\pi}2}{\dfrac{4\pi}3}=-\dfrac3{4\pi},$$ $$f_2(x)>\dfrac{4\pi-3}{8\pi }.\tag7$$ Therefore, inequality $$(4)$$ is proved.

$$\color{brown}{\textbf{Done!}}$$

• @hd_30102 I am glad too Commented Feb 13, 2021 at 4:27